HDU-2602-Bone Collector（01背包）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602

Bone Collector

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

 

题目很简单，给出石头的数量和背包的最大容量，每块石头的价值、体积，单纯的01背包问题。
代码如下：

#include<iostream>#include<algorithm>#include<cstring>using namespace std;int main(){int T,n,m,value[1001],cost[1001],dp[1001];cin>>T;while(T--){cin>>n>>m;int i,j;memset(value,0,sizeof(value));memset(cost,0,sizeof(cost));memset(dp,0,sizeof(dp));for(i=1;i<=n;i++)cin>>value[i];for(i=1;i<=n;i++)cin>>cost[i];for(i=1;i<=n;i++){for(j=m;j>=cost[i];j--){dp[j]=max(dp[j],dp[j-cost[i]]+value[i]);}}cout<<dp[m]<<endl;}return 0;}