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krahets
2024-04-09 20:43:40 +08:00
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9
docs/chapter_dynamic_programming/knapsack_problem.md
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@ -317,7 +317,7 @@ $$
val no = knapsackDFS(wgt, value, i - 1, c)
val yes = knapsackDFS(wgt, value, i - 1, c - wgt[i - 1]) + value[i - 1]
// 返回两种方案中价值更大的那一个
return max(no.toDouble(), yes.toDouble()).toInt()
return max(no, yes)
}
```
@ -692,7 +692,7 @@ $$
val no = knapsackDFSMem(wgt, value, mem, i - 1, c)
val yes = knapsackDFSMem(wgt, value, mem, i - 1, c - wgt[i - 1]) + value[i - 1]
// 记录并返回两种方案中价值更大的那一个
mem[i][c] = max(no.toDouble(), yes.toDouble()).toInt()
mem[i][c] = max(no, yes)
return mem[i][c]
}
```
@ -1052,8 +1052,7 @@ $$
dp[i][c] = dp[i - 1][c]
} else {
// 不选和选物品 i 这两种方案的较大值
dp[i][c] = max(dp[i - 1][c].toDouble(), (dp[i - 1][c - wgt[i - 1]] + value[i - 1]).toDouble())
.toInt()
dp[i][c] = max(dp[i - 1][c], dp[i - 1][c - wgt[i - 1]] + value[i - 1])
}
}
}
@ -1445,7 +1444,7 @@ $$
if (wgt[i - 1] <= c) {
// 不选和选物品 i 这两种方案的较大值
dp[c] =
max(dp[c].toDouble(), (dp[c - wgt[i - 1]] + value[i - 1]).toDouble()).toInt()
max(dp[c], dp[c - wgt[i - 1]] + value[i - 1])
}
}
}