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=;i<=n;i++ )
scanf("%d",&worth[i]);
for( int i=;i<=n;i++ )
scanf("%d",&cost[i]);
memset(dp,,sizeof(dp));
for( int i=;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 ;
}