Bone Collector（基础 0-1背包）

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 2³¹).

 

Sample Input

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14
#include<stdio.h>#include<stdlib.h>#include<string.h>int max(int a, int b){    return a>b?a:b;}int main(){    int t,j,i,dv;    int n;    int dp[1005][1005],gv[1005],gz[1005];    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof(dp));        memset(gv,0,sizeof(gv));        memset(gz,0,sizeof(gz));        scanf("%d%d",&n,&dv);        for(i=1;i<=n;i++)            scanf("%d",&gz[i]); //从1开始 因为下面要用到i-1。        for(i=1;i<=n;i++)            scanf("%d",&gv[i]); //从1开始 因为下面要用到i-1。        for(i=1;i<=n;i++)            for(j=0;j<=dv;j++)            {                if(j<gv[i])                    dp[i][j]=dp[i-1][j]; //注意此处是i-1。                else                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-gv[i]]+gz[i]);//注意此处是i-1。             }        printf("%d\n",dp[n][dv]);    }    return 0;}