M - 基础DP

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


题目大意：有N个数，背包的容量为V，第一行为价值，第二行为对应的体积。那么题目就简单了。著名的背包问题
for(i=0;i<n;i++)
          for(j=v;j>=volum[i];j--)
               bb[j]=max(bb[j],bb[j-volum[i]]+value[i]);//模板哟

代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int main()
{    int value[1000],volum[1000],bb[1000],n,v,t,i,j;
     scanf("%d",&t);
     while(t--)
     {
          scanf("%d%d",&n,&v);
          for(i=0;i<n;i++)
          scanf("%d",&value[i]);
          for(i=0;i<n;i++)
          scanf("%d",&volum[i]);
          memset(bb,0,sizeof(bb));
          for(i=0;i<n;i++)
          for(j=v;j>=volum[i];j--)
               bb[j]=max(bb[j],bb[j-volum[i]]+value[i]);//不断更新每个体积可以放的最大价值。
          printf("%d\n",bb[v]);
     }
     return 0;
}