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  • HDU2602-Bone Collector

    描述:

      Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave … 

      The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ? 

     

      The first line contain a integer T , the number of cases. 

      Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

      One integer per line representing the maximum of the total value (this number will be less than 2 31).

    代码:

      最基本的01背包。

    #include<stdio.h>
    #include<string.h>
    #include<iostream>
    #include<stdlib.h>
    #include <math.h>
    using namespace std;
    #define N 1005
    
    int main(){
        int T,n,v;
        int worth[N],cost[N],dp[N];
        scanf("%d",&T);
        while( T-- ){
            scanf("%d%d",&n,&v);
            for( int i=1;i<=n;i++ )
                scanf("%d",&worth[i]);
            for( int i=1;i<=n;i++ )
                scanf("%d",&cost[i]);
            memset(dp,0,sizeof(dp));
            for( int i=1;i<=n;i++ ){
                for( int j=v;j>=cost[i];j-- ){
                    dp[j]=max(dp[j],dp[j-cost[i]]+worth[i]);
                }
            }
            printf("%d
    ",dp[v]);
        }
        system("pause");
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/lucio_yz/p/4748942.html
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