5.分组背包问题

 

 分组背包问题是物品有n组，每组里面有若干个物品，每组里最多选一个。

dp[i][j]只从前i组物品中选，且总体积不大于j的最大价值

状态计算：枚举第i组物品选哪个或不选

不选第i组物品，选第i组的第1个物品，选第i组的第2个物品，等等等等

 二维做法

#include <bits/stdc++.h>
using namespace std;
const int N = 110; //一共有100组物品 
int s[N]; //每组物品所含的物品数 
int v[N][N], w[N][N];
int dp[N][N];
int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        cin >> s[i];
        for (int j = 1; j <= s[i]; j++) {
            cin >> v[i][j] >> w[i][j];
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
            dp[i][j] = dp[i - 1][j]; //不选 
            for (int k = 1; k <= s[i]; k++) {
                if (j >= v[i][k]) {
                    dp[i][j] = max(dp[i][j], dp[i - 1][j - v[i][k]] + w[i][k]);
                }
            }
        }
    }
    cout << dp[n][m] << endl;
    return 0;
}

一维做法

#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int s[N], v[N][N], w[N][N], dp[N];
int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        cin >> s[i];
        for (int j = 1; j <= s[i]; j++) {
            cin >> v[i][j] >> w[i][j];
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = m; j >= 0; j--) {
            for (int k = 1; k <= s[i]; k++) {
                if (j >= v[i][k]) {
                    dp[j] = max(dp[j], dp[j - v[i][k]] + w[i][k]);
                }
            }
        }
    }
    cout << dp[m] << endl;
    return 0;
}