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@ -87,48 +87,30 @@ $f(i,j,k)$ 表示从前$i$个物品中选,且花费$1$**不少于**$j$,花费$
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**答**:不需要,都可以满足要求了,即$j=0,k=0$,也就是$f[i-1][0][0]+w$,而对于一个无欲无求的$f[i-1][0][0]$自然是等于$0$,也就是$f[i][j][k]=w$
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### 三、三维朴素解法
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### 三、三维解法
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 1010;
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const int M = 110;
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int n, m, m1, m2;
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int f[N][M][M];
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int n, m1, m2;
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//二维费用01背包-不少于维度费用,求最小代价
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int main() {
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//注意次序
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scanf("%d %d %d", &m1, &m2, &n);
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//求最小值
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cin >> m1 >> m2 >> n;
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memset(f, 0x3f, sizeof f);
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f[0][0][0] = 0;
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for (int i = 1; i <= n; i++) {
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int v1, v2, w;
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scanf("%d %d %d", &v1, &v2, &w);
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cin >> v1 >> v2 >> w;
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for (int j = 0; j <= m1; j++)
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for (int k = 0; k <= m2; k++) {
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//不选择i号物品
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f[i][j][k] = f[i - 1][j][k];
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//分情况讨论
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//物品i加上就够一维使用,此时,只关心二维情况即可
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if (j - v1 < 0 && k - v2 >= 0)
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f[i][j][k] = min(f[i][j][k], f[i - 1][0][k - v2] + w);
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//物品i加上就够二维使用,此时,只关心一维情况即可
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else if (j - v1 >= 0 && k - v2 < 0)
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f[i][j][k] = min(f[i][j][k], f[i - 1][j - v1][0] + w);
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//如果选择了i号物品,两个维度直接拉满,那么只选择一个i就足够用,它参选的价值是w
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else if (j - v1 < 0 && k - v2 < 0)
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f[i][j][k] = min(f[i][j][k], w);
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else
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//正常递推
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f[i][j][k] = min(f[i][j][k], f[i - 1][j - v1][k - v2] + w);
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f[i][j][k] = min(f[i - 1][j][k], f[i - 1][max(0, j - v1)][max(0, k - v2)] + w);
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}
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}
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printf("%d\n", f[n][m1][m2]);
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cout << f[n][m1][m2] << endl;
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return 0;
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}
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```
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