HDU2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 52183    Accepted Submission(s): 21988

Problem 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 2³¹).

 

Sample Input

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14
0-1背包练手
容量为V的袋子，每个骨头有其体积和价值
问最大价值为多少。
动态转移方程：
if (dp[j-vol[i]]+val[i]>dp[j]) dp[j]=dp[j-vol[i]]+val[i];

#include <cstdio>#include <string.h>const int MAX=1005;int dp[1005],val[MAX],vol[MAX];int main(){    int T,n,v,i,j;    scanf("%d",&T);    while(T--)    {        memset(dp,0,sizeof(dp));        scanf("%d%d",&n,&v);        for(i=1;i<=n;i++) scanf("%d",&val[i]);        for(i=1;i<=n;i++) scanf("%d",&vol[i]);        for(i=1;i<=n;i++)            for(j=v;j>=vol[i];j--)                if (dp[j-vol[i]]+val[i]>dp[j]) dp[j]=dp[j-vol[i]]+val[i];        printf("%d\n",dp[v]);    }    return 0;}