hdu2602 Bone Collector(背包问题)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 39612    Accepted Submission(s): 16412

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

 

Author

Teddy

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

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非常常规的0 1背包问题。

第一排输入的是价值，第二排是体积

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