版本一： 时间复杂度$O(nm^2)$, 所以TLE

#include<iostream>

using namespace std;
const int N = 1010;
int n, m;
int f[N][N];
int v[N], w[N];

int main() {
    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 = 0; j <= m; j++) {
            for (int k = 0; k * v[i] <= j; k++) {
                f[i][j] = max(f[i][j], f[i - 1][j - v[i] * k] + w[i] * k);
            }
        }
    }
    cout << f[n][m] << endl;
}

版本二 一维优化

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1010;

int n, m;
int v[N], w[N];
int f[N];

int main() {
    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 = v[i]; j <= m; j++)
            f[j] = max(f[j], f[j - v[i]] + w[i]);

    cout << f[m] << endl;

    return 0;
}

01 背包问题 优化成一维 j维度从大到小枚举
f[i][j]=max(f[i][j], f[i-1][j-v[i]+w[i]);

完全背包问题 优化成一维 j维度从小到大枚举
f[i][j]=max(f[i][j], f[i][j-v[i]+w[i]);