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 1005int 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\n",dp[v]);}system("pause");return 0; }

 

转载于:https://www.cnblogs.com/lucio_yz/p/4748942.html

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