小班的储钱罐

Posted 2020-12-24 dai-jia-ye

裸地DP，方程为：dp[j]=min(dp[j],dp[j-b[i]]+a[i])。代码：

#include<iostream>
#include<cmath>
#include<cstring>
#define INF 0x3f3f3f3f
using namespace std;
int n,m,k;
int dp[100001],a[110000],b[1100000];
int main()
{
    cin>>n>>m;
    cin>>k;
    m=m-n;//剩余空间 
    for(int i=1;i<=k;i++)
    {
        cin>>a[i]>>b[i];
    }
    memset(dp,INF,sizeof(dp));
    dp[0]=0;
    for(int i=1;i<=k;i++)
    {
        for(int j=b[i];j<=m;j++)//从b[i]开始 
        {
            dp[j]=min(dp[j],dp[j-b[i]]+a[i]);
        }
    }
    if(dp[m]!=INF)
    {
        cout<<"The minimum amount of money in the piggy-bank is "<<dp[m]<<".";
    }
    else
    {
        cout<<"This is impossible.";
    }
}

　　嗯~~