hdu 1114 Piggy-Bank

题目:

链接:点击打开链接

题意:

知道存钱罐的质量和装满硬币的存钱罐的质量,然后是不同硬币的价值和质量,求出存钱罐里钱币的最小价值。

算法:

完全背包问题,银币的个数是不限的。

思路:

状态转移方程:j = 0时,价值为0

dp[j] = min(dp[j],dp[j-w[i]]+v[i]);//表示质量为j的钱币,含有的最小的价值

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define INF 10000000
int dp[10010],w[555],v[555];
int e,f;

int main()
{
    //freopen("input.txt","r",stdin);
    int t,coinw,n;
    cin>>t;
    while(t--)
    {
        cin>>e>>f;
        coinw = f-e;
        cin>>n;
        for(int i=0; i<=coinw; i++)
            dp[i] = INF;
        dp[0] = 0;
        for(int i=0; i<n; i++)
        {
            cin>>v[i]>>w[i];
        }
        for(int i=0; i<n; i++)
        {
            for(int j=w[i]; j<=coinw; j++)
            {
                dp[j] = min(dp[j],dp[j-w[i]]+v[i]);
            }
        }
        if(dp[coinw] == INF)
            printf("This is impossible.\n");
        else
            printf("The minimum amount of money in the piggy-bank is %d.\n",dp[coinw]);
    }
    return 0;
}

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时间: 05-16

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