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;
}