求背包容量恰好被装满时,最大价值为多少?
需要将数组设置为很小的负数,这样表示状态不可达
设置f[0] = 0表示背包容量为0是物体价值为0,初始化
#include <algorithm>
#include <cstring>
#include <iostream>
using namespace std;
const int N = 1010;
int v[N], w[N], f[N], g[N];
int main() {
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; i ++) {
cin >> v[i] >> w[i];
}
for (int i = 1; i <= n; i ++) {
for (int j = m; j >= v[i]; j --) {
f[j] = max(f[j], f[j - v[i]] + w[i]);
}
}
cout << f[m] << endl;
memset(f, -0x3f, sizeof f);
f[0] = 0;
for (int i = 1; i <= n; i ++) {
for (int j = m; j >= v[i]; j --) {
if(f[j - v[i]] != 0x3f) f[j] = max(f[j], f[j - v[i]] + w[i]);
}
}
if(f[m] < 0) f[m] = 0;
cout << f[m];
return 0;
}

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