mirror of
https://github.com/youngyangyang04/leetcode-master.git
synced 2025-07-07 07:35:35 +08:00
Update
This commit is contained in:
@ -613,4 +613,3 @@ func main() {
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}
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```
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<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
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|
@ -927,4 +927,3 @@ func main() {
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### C
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|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
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|
@ -942,4 +942,3 @@ main()
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### C
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|
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<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
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|
@ -760,4 +760,3 @@ int main()
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}
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```
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<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
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|
@ -757,4 +757,3 @@ main()
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### C
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<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
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|
@ -432,4 +432,3 @@ echo $s;
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### Rust:
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<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
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|
@ -409,4 +409,3 @@ echo $s;
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### Rust:
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<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
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|
@ -408,4 +408,3 @@ func main() {
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}
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```
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<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
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|
@ -536,4 +536,3 @@ main()
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<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
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|
@ -538,4 +538,3 @@ main()
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### C
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<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
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@ -183,7 +183,6 @@ struct Edge { //邻接表
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Edge(int t, int w): to(t), val(w) {} // 构造函数
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};
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int main() {
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int n, m, p1, p2, val;
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cin >> n >> m;
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@ -229,6 +228,103 @@ int main() {
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}
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}
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}
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if (flag) cout << "circle" << endl;
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else if (minDist[end] == INT_MAX) {
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cout << "unconnected" << endl;
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} else {
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cout << minDist[end] << endl;
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}
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}
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```
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以上代码看上去没问题,但提交之后会发现报错了,为什么呢?
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例如这组数据:
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```
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4 6
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1 4 3
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1 2 1
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1 3 1
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3 2 -2
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2 4 1
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3 4 0
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```
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如图:
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上面代码在执行的过程中,节点4 已经出队列了,但 计算入度会反复计算 节点4的入度,导致 误判 该图有环。
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我们需要记录哪些节点已经出队列了,哪些节点在队列里面,对于已经出队列的节点不用统计入度,以下为优化后的代码:
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```CPP
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#include <iostream>
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#include <vector>
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#include <queue>
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#include <list>
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#include <climits>
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using namespace std;
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struct Edge { //邻接表
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int to; // 链接的节点
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int val; // 边的权重
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Edge(int t, int w): to(t), val(w) {} // 构造函数
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};
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int main() {
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int n, m, p1, p2, val;
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cin >> n >> m;
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vector<list<Edge>> grid(n + 1); // 邻接表
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// 将所有边保存起来
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for(int i = 0; i < m; i++){
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cin >> p1 >> p2 >> val;
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// p1 指向 p2,权值为 val
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grid[p1].push_back(Edge(p2, val));
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}
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int start = 1; // 起点
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int end = n; // 终点
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vector<int> minDist(n + 1 , INT_MAX);
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minDist[start] = 0;
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queue<int> que;
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que.push(start); // 队列里放入起点
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vector<int> count(n+1, 0); // 记录节点加入队列几次
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count[start]++;
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vector<bool> inQueue(n+1, false); // 记录节点是否在队列中
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bool flag = false;
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while (!que.empty()) {
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int node = que.front(); que.pop();
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inQueue[node] = false; // 节点出队
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for (Edge edge : grid[node]) {
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int from = node;
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int to = edge.to;
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int value = edge.val;
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if (minDist[to] > minDist[from] + value) { // 开始松弛
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minDist[to] = minDist[from] + value;
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if (!inQueue[to]) { // 避免重复入队
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que.push(to);
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inQueue[to] = true;
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count[to]++;
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if (count[to] == n) {// 如果加入队列次数超过 n-1次 就说明该图与负权回路
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flag = true;
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while (!que.empty()) que.pop();
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break;
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}
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}
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}
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}
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}
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if (flag) cout << "circle" << endl;
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else if (minDist[end] == INT_MAX) {
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@ -236,9 +332,7 @@ int main() {
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} else {
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cout << minDist[end] << endl;
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}
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}
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```
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## 其他语言版本
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@ -456,4 +550,3 @@ if __name__ == "__main__":
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### C
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||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
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|
@ -922,4 +922,3 @@ if __name__ == "__main__":
|
||||
|
||||
### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
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|
@ -574,4 +574,3 @@ if __name__ == '__main__':
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|
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### C
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||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
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|
@ -72,6 +72,7 @@
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* 1 <= N <= 100
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* 1 <= M <= 500
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[所有可达路径](https://www.bilibili.com/video/BV1VePeepEpP),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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## 插曲
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@ -887,4 +888,3 @@ async function dfs(graph, x, n) {
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||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
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|
@ -46,6 +46,8 @@
|
||||
|
||||
## 思路
|
||||
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[图论:广搜有陷阱啊!! | 广度优先搜索 | 卡码网:99.岛屿数量](https://www.bilibili.com/video/BV18PRGYcEiB/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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注意题目中每座岛屿只能由**水平方向和/或竖直方向上**相邻的陆地连接形成。
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也就是说斜角度链接是不算了, 例如示例二,是三个岛屿,如图:
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@ -557,4 +559,3 @@ object Solution {
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### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -46,6 +46,8 @@
|
||||
|
||||
## 思路
|
||||
|
||||
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[图论:来用深搜解决一道题目,两种深搜写法,你掉坑了吗? | 卡码网:99.岛屿数量](https://www.bilibili.com/video/BV18PRGYcEiB/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
|
||||
|
||||
注意题目中每座岛屿只能由**水平方向和/或竖直方向上**相邻的陆地连接形成。
|
||||
|
||||
也就是说斜角度链接是不算了, 例如示例二,是三个岛屿,如图:
|
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@ -459,4 +461,3 @@ object Solution {
|
||||
|
||||
### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -44,6 +44,8 @@
|
||||
|
||||
## 思路
|
||||
|
||||
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[图论:深搜也有陷阱啊!! | 深搜优先搜索 | 卡码网:100.岛屿的最大面积](https://www.bilibili.com/video/BV1FzoyY5EXH),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
|
||||
|
||||
注意题目中每座岛屿只能由**水平方向和/或竖直方向上**相邻的陆地连接形成。
|
||||
|
||||
也就是说斜角度链接是不算了, 例如示例二,是三个岛屿,如图:
|
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@ -890,4 +892,3 @@ main();
|
||||
|
||||
### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -48,6 +48,8 @@
|
||||
|
||||
## 思路
|
||||
|
||||
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[图论:岛屿问题再出新花样 | 深搜优先搜索 | 卡码网:101.孤岛总面积](https://www.bilibili.com/video/BV1mmZJYRESc),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
|
||||
|
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本题使用dfs,bfs,并查集都是可以的。
|
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|
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本题要求找到不靠边的陆地面积,那么我们只要从周边找到陆地然后 通过 dfs或者bfs 将周边靠陆地且相邻的陆地都变成海洋,然后再去重新遍历地图 统计此时还剩下的陆地就可以了。
|
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@ -700,4 +702,3 @@ const bfs = (graph, x, y) => {
|
||||
|
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### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -55,6 +55,8 @@
|
||||
|
||||
## 思路
|
||||
|
||||
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[图论:岛屿问题也需要逆向思考 | 深搜优先搜索DFS | 卡码网:102.沉没孤岛](https://www.bilibili.com/video/BV1fjdWYyEGu),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
|
||||
|
||||
这道题目和[0101.孤岛的总面积](https://kamacoder.com/problempage.php?pid=1173)正好反过来了,[0101.孤岛的总面积](https://kamacoder.com/problempage.php?pid=1173)是求 地图中间的空格数,而本题是要把地图中间的 1 都改成 0 。
|
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|
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那么两题在思路上也是差不多的。
|
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@ -503,4 +505,3 @@ const bfs = (graph, x, y) => {
|
||||
|
||||
### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -59,6 +59,8 @@
|
||||
|
||||
## 思路
|
||||
|
||||
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[图论:水流向何方? | 深搜优先搜索DFS| 广度优先搜索BFS | 卡码网:103.水流问题](https://www.bilibili.com/video/BV1WNoEYrEio),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
|
||||
|
||||
一个比较直白的想法,其实就是 遍历每个点,然后看这个点 能不能同时到达第一组边界和第二组边界。
|
||||
|
||||
至于遍历方式,可以用dfs,也可以用bfs,以下用dfs来举例。
|
||||
@ -854,4 +856,3 @@ const isResult = (x, y) => {
|
||||
|
||||
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -50,6 +50,8 @@
|
||||
|
||||
## 思路
|
||||
|
||||
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[图论:岛屿问题上难度了! |深搜优先搜索DFS | 广度优先搜索BFS | 卡码网:104.建造最大岛屿](https://www.bilibili.com/video/BV1Dn5CzZEw1/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
|
||||
|
||||
本题的一个暴力想法,应该是遍历地图尝试 将每一个 0 改成1,然后去搜索地图中的最大的岛屿面积。
|
||||
|
||||
计算地图的最大面积:遍历地图 + 深搜岛屿,时间复杂度为 n * n。
|
||||
@ -663,4 +665,3 @@ const dfs = (graph, visited, x, y, mark) => {
|
||||
|
||||
### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -1,7 +1,7 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们受益!</strong></p>
|
||||
|
||||
# 105.有向图的完全可达性
|
||||
# 105.有向图的完全联通
|
||||
|
||||
[卡码网题目链接(ACM模式)](https://kamacoder.com/problempage.php?pid=1177)
|
||||
|
||||
@ -44,6 +44,8 @@
|
||||
|
||||
## 思路
|
||||
|
||||
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[图论:这次我们能到达有向图的所有节点吗?| 深度优先搜索 | 卡码网:105. 有向图的完全联通](https://www.bilibili.com/video/BV1QRJ6ziEvJ),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
|
||||
|
||||
本题给我们是一个有向图, 意识到这是有向图很重要!
|
||||
|
||||
接下来我们再画一个图,从图里可以直观看出来,节点6 是 不能到达节点1 的
|
||||
@ -553,4 +555,3 @@ rl.on('close',()=>{
|
||||
|
||||
### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -47,6 +47,8 @@
|
||||
|
||||
## 思路
|
||||
|
||||
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[图论:这次的岛屿问题有点简单了。。 卡码网:106.岛屿的周长](https://www.bilibili.com/video/BV13fjczSEyV),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
|
||||
|
||||
岛屿问题最容易让人想到BFS或者DFS,但本题确实还用不上。
|
||||
|
||||
为了避免大家惯性思维,所以给大家安排了这道题目。
|
||||
@ -359,4 +361,3 @@ func parseLine(line string, count int) []int {
|
||||
|
||||
### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -422,4 +422,3 @@ const isSame = (u, v) => {
|
||||
|
||||
### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -376,4 +376,3 @@ const isSame = (u, v) => {
|
||||
|
||||
### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -596,4 +596,3 @@ const getRemoveEdge = (edges) => {
|
||||
|
||||
### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -63,6 +63,8 @@ yhn
|
||||
|
||||
## 思路
|
||||
|
||||
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[图论:朴实无华的广搜这么难? | 广度优先搜索 | 卡码网:110.字符串接龙](https://www.bilibili.com/video/BV1QEEizDEC4),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
|
||||
|
||||
以示例1为例,从这个图中可以看出 abc 到 def的路线 不止一条,但最短的一条路径上是4个节点。
|
||||
|
||||

|
||||
@ -360,4 +362,3 @@ const init = async () => {
|
||||
|
||||
### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -539,4 +539,3 @@ const init = async () => {
|
||||
|
||||
### C
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -684,4 +684,3 @@ int main() {
|
||||
|
||||
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -91,4 +91,3 @@ cout << result[result.size() - 1];
|
||||
等大家将图论刷完,就会感受到我的良苦用心。加油
|
||||
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -454,4 +454,3 @@ void join(int u, int v) {
|
||||
|
||||
敬请期待 并查集题目精讲系列。
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -1,5 +1,7 @@
|
||||
# 广度优先搜索理论基础
|
||||
|
||||
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[广度优先搜索理论基础](https://www.bilibili.com/video/BV1M19iY4EL9),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
|
||||
|
||||
在[深度优先搜索](./图论深搜理论基础.md)的讲解中,我们就讲过深度优先搜索和广度优先搜索的区别。
|
||||
|
||||
广搜(bfs)是一圈一圈的搜索过程,和深搜(dfs)是一条路跑到黑然后再回溯。
|
||||
@ -102,4 +104,3 @@ void bfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y
|
||||
|
||||
相信看完本篇,大家会对广搜有一个基础性的认识,后面再来做对应的题目就会得心应手一些。
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -143,4 +143,3 @@ kruscal的主要思路:
|
||||
图论也是我 《代码随想录》所有章节里 所费精力最大的一个章节。
|
||||
|
||||
只为了不负录友们的期待。 大家加油💪🏻
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -1,6 +1,8 @@
|
||||
|
||||
# 深度优先搜索理论基础
|
||||
|
||||
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[深度优先搜索理论基础](https://www.bilibili.com/video/BV1hFA8eKE6C),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
|
||||
|
||||
录友们期待图论内容已久了,为什么鸽了这么久,主要是最近半年开始更新[代码随想录算法公开课](https://www.bilibili.com/video/BV1fA4y1o715/),是开源在B站的算法视频,已经帮助非常多基础不好的录友学习算法。
|
||||
|
||||
录视频其实是非常累的,也要花很多时间,所以图论这边就没抽出时间来。
|
||||
@ -194,4 +196,3 @@ for (选择:本节点所连接的其他节点) {
|
||||
后面我也会给大家安排具体练习的题目,依旧是代码随想录的风格,循序渐进由浅入深!
|
||||
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -1,6 +1,8 @@
|
||||
|
||||
# 图论理论基础
|
||||
|
||||
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[图论理论基础](https://www.bilibili.com/video/BV1hYNBeYEvb/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
|
||||
|
||||
这一篇我们正式开始图论!
|
||||
|
||||
代码随想录图论中的算法题目将统一使用ACM模式,[为什么要使用ACM模式](./图论为什么用ACM模式.md)
|
||||
@ -246,4 +248,3 @@ dfs 和 bfs 一种搜索算法,可以在不同的数据结构上进行搜索
|
||||
敬请期待!
|
||||
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
@ -51,4 +51,3 @@
|
||||
|
||||
|
||||
|
||||
<div align="center"><a href="https://www.nowcoder.com/link/wx_sqlcarl199" target="_blank"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220416223343.png" width="1000"/></a></div>
|
||||
|
Reference in New Issue
Block a user