HDU 1114 Piggy-Bank

该题利用了我们的逆向思维，同时要注意该题他的质量是一定的，也就是说背包一定要是满的，刚开始对于这类背包我们令初始值是负无穷，而这题则相反，令初始值是正无穷，每次区最小的数，同时要注意f[j-weight[i]]!=inf应为一相等就与背包一定要满的条件相矛盾；

#include<stdio.h>
#include<stdlib.h>
const int inf=0x7fffffff;
int DP( int weight[],int val[],int  N,int vol )
{
    int f[50024]={0};
    for( int i=1; i<=vol; i++ )
       f[i]=inf;
    for( int i=1; i<=N; i++ )
    {
        for( int j=0; j<=vol; j++ )
           if( weight[i]<=j&&f[j-weight[i]]!=inf&&f[j-weight[i]]+val[i]<f[j] )
                 f[j]=f[j-weight[i]]+val[i];     
    }
    return f[vol];    
}
int main()
{
    int N,n,weight[524],val[524],E,F;
    scanf( "%d",&n );
    for( int i=0; i<n;i++ )
    {
        scanf( "%d%d",&E,&F );
        scanf( "%d",&N );
        for( int j=1; j<=N; j++ )
          scanf( "%d%d",&val[j],&weight[j] );
          int t=DP( weight, val,N,F-E );
        if( t==inf )
        {
            printf( "This is impossible.\n" );    
        }
        else printf( "The minimum amount of money in the piggy-bank is %d.\n",t );     
    }
    return 0;    
}