Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 81450 Accepted Submission(s): 33681

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

5 10

1 2 3 4 5

5 4 3 2 1

Sample Output

解题思路：

典型的01背包。背包体积为10，有5件物品，价值分别为1 2 3 4 5，体积分别为5 4 3 2 1，求最大价值。

代码：

#include<bits/stdc++.h>
using namespace std;
int w[1005],v[1005],dp[1005];
int main()
{
	int repeat;
	scanf("%d",&repeat);
	while(repeat--){
		int N,V;
		scanf("%d%d",&N,&V);
		memset(w,0,sizeof(w));
		memset(v,0,sizeof(v));
		memset(dp,0,sizeof(dp));
		for(int i=0;i<N;i++)scanf("%d",&v[i]);
		for(int i=0;i<N;i++)scanf("%d",&w[i]);
		for(int i=0;i<N;i++){
			for(int j=V;j>=w[i];j--){
				dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
			}
		}
		printf("%d\n",dp[V]);
	}
	return 0;
 }

【HDU2602】Bone Collector（01背包模板题）

Bone Collector

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