完全背包问题_找的是最小值

完全背包问题_找的是最小值

Sample Input

3 10 110 2 1 1 30 50 10 110 2 1 1 50 30 1 6 2 10 3 20 4

Sample Output

The minimum amount of money in the piggy-bank is 60. The minimum amount of money in the piggy-bank is 100. This is impossible.

这题是完全背包问题，求的是最小值，那么就要在初始化时将临时数组f赋值为最大值，并f[0]=0;这样改变下循环判断条件就好了，最后如果不能完全匹配，也就是f[最大值]不变，便输出不可能。

#include<iostream>
using namespace std;
struct good
{
    int c,w;
}goods[501];
int f[10009];
int main()
{
    int t,e,ff,s,n,i,j;
    cin>>t;
    while(t--)
    {
        cin>>e>>ff;
        s=ff-e;
        for(i=0;i<=s;i++)f[i]=1000000000;
        f[0]=0;
        cin>>n;
        for(i=0;i<n;i++)cin>>goods[i].w>>goods[i].c;
        for(i=0;i<n;i++)
        {
            for(j=goods[i].c;j<=s;j++)
            {
                if(f[j]>f[j-goods[i].c]+goods[i].w)
                {
                    f[j]=f[j-goods[i].c]+goods[i].w;
                }
            }
        }
        if(f[s]==1000000000)cout<<"This is impossible."<<endl;
        else cout<<"The minimum amount of money in the piggy-bank is "<<f[s]<<"."<<endl;
    }
    return 0;
}