Bone Collector （简单dp）

最新推荐文章于 2020-04-02 16:55:44 发布

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最新推荐文章于 2020-04-02 16:55:44 发布

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分类专栏： # 基础DP(动态规划)

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本文介绍了一种经典的背包问题——骨收集者问题。该问题是关于一名骨收集者如何在有限背包容量下，从一系列不同价值与体积的骨头中选择最优组合以最大化总价值。文章通过给出具体的输入输出示例及C++实现代码，详细阐述了如何解决此类问题。

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Bone Collector

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 2 ³¹).

Sample Input

Sample Output

code：

#include <iostream>
#include <cstring>
using namespace std;
int main(){
	int t;
	cin >> t;
	while(t--){
		int n,W;
		cin >> n >> W;
		long dp[1002];
		long v[1002],w[1002];
		int i,j;
		memset(dp,0,sizeof(dp));
		for(i = 1; i <= n; i++)
		  cin >> v[i];
		for(i = 1; i <= n; i++)
		  cin >> w[i];
		for(i = 1; i <= n; i++){
			for(j = W; j >= w[i]; j--){
				dp[j] = max(dp[j],dp[j-w[i]]+v[i]);
			}
		}
		cout << dp[W] << endl;
	}
	return 0;
}