多重背包问题

输入样例

4 5
1 2 3
2 4 1
3 4 3
4 5 2

输入样例

10

朴素算法

代码

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;

const int N = 110;
int n, m;
int v[N], w[N], s[N];
int f[N][N];

//f[i][j]表示只从前i个物品中选,并且总体积小于j
//状态转移方程和完全背包类似

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        scanf("%d%d%d", &v[i], &w[i], &s[i]);

    for (int i = 1; i <= n; i++)
        for (int j = 0; j <= m; j++)
            for (int k = 0; k <= s[i] && k * v[i] <= j; k++)//表示物品的个数
                f[i][j] = max(f[i][j], f[i - 1][j - k * v[i]] + k * w[i]);

    cout << f[n][m];
    return 0;
}

二进制倍增优化

代码

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;

const int N = 25000;
int n, m;
int v[N], w[N];
int f[N];

int main()
{
    cin >> n >> m;

    int cnt = 0;//表示物品个数
    for (int i = 1; i <= n; i++)
    {
        int a, b, s;
        cin >> a >> b >> s;

        int k = 1;
        while (k <= s)
        {
            cnt++;
            v[cnt] = a * k;//k个物品的体积
            w[cnt] = b * k;//k个物品的权重
            s -= k;
            k *= 2;//物品个数每次增加2倍
        }
        if (s > 0)//剩余物品看做一组
        {
            cnt++;
            v[cnt] = a * s;
            w[cnt] = b * s;
        }
    }

    n = cnt;
    for (int i = 1; i <= n; i++)//转化为01背包问题
        for (int j = m; j >= v[i]; j--)
            f[j] = max(f[j], f[j - v[i]] + w[i]);

    cout << f[m];
    return 0;
}