HDU 2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 45816    Accepted Submission(s): 19072

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

 

简单01背包问题，这些都一个解法，用一维的效率好很多。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int MAXN=1005;const int MAXV=1005;int value[MAXN],weight[MAXN];int dp[MAXV];int main(){        int t;        scanf("%d",&t);        while(t--)        {                int N,V;                scanf("%d%d",&N,&V);                memset(dp,0,sizeof(dp));                for(int i=0;i<N;i++)                        scanf("%d",&value[i]);                for(int i=0;i<N;i++)                        scanf("%d",&weight[i]);                for(int i=0;i<N;i++)                        for(int j=V;j>=weight[i];j--)                                dp[j]=max(dp[j],dp[j-weight[i]]+value[i]);                printf("%d\n",dp[V]);        }        return 0;}