hdu 2602

Description

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 ?



 

Input

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.
 

Output

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

Sample Input

1
5 10
1 2 3 4 5
5 4 3 2 1

 

Sample Output

14 题意：第一行输入案例数，第二行输入两个整数N,V，分别代表骨头的数量和袋子的体积，接下来的两行分别有N个整数，分别代表骨头的价值和体积，问你在袋子装满的情况下价值最大，求出最大价值。思路：典型的0-1背包问题，让价值作为背包，value[i]代表骨头的体积，volume[i]代表骨头的体积，dp[i]代表骨头的价值，求出其最大值就好了。代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
int value[1010];//价值
int volume[1010];//体积
int dp[1010];//价值作为背包
using namespace std;
int main()
{
int T;
int N,V;
int sum;
scanf("%d",&T);
while(T--)
{
sum=0;
scanf("%d%d",&N,&V);
for(int i=0; i<N; i++)
scanf("%d",&value[i]);                //输入
for(int i=0; i<N; i++)
{
scanf("%d",&volume[i]);
//sum+=volume[i];
}
memset(dp,0,sizeof(dp));
//      dp[0]=0;//价值为0的时候肯定为V
for(int i=0; i<N; i++)
for(int j=V; j>=volume[i]; j--)
dp[j]=max(dp[j],dp[j-volume[i]]+value[i]);

printf("%d\n",dp[V]);

}
return 0;
}