HDU 2602 Bone Collector 最简单背包

最新推荐文章于 2022-10-28 13:05:08 发布

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最新推荐文章于 2022-10-28 13:05:08 发布

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分类专栏：动态规划水题文章标签： HDU 水题

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本文介绍了一个经典的背包问题案例，通过编程解决如何在限定体积内获取最大价值的问题。使用了动态规划的方法来解决此问题，并提供了完整的C++代码实现。

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 65995 Accepted Submission(s): 27523

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

Sample Output

这个真的是一点弯弯都没有，所以就直接代码走起来.

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define N 1005

using namespace std;

int n,m;
int dp[N],w[N],v[N];

int main()
{
	int ccase,i,j;
	scanf("%d",&ccase);
	while(ccase--)
	{
		memset(dp,0,sizeof(dp));
		memset(w,0,sizeof(w));
		memset(v,0,sizeof(v));
		scanf("%d %d",&n,&m);
		for(i=1;i<=n;i++) scanf("%d",&v[i]);
		for(i=1;i<=n;i++) scanf("%d",&w[i]);
		for(i=1;i<=n;i++)
		{
			for(j=m;j>=w[i];j--)
			{
				dp[j] = dp[j] > dp[j - w[i]] + v[i] ? dp[j] : dp[j - w[i]] + v[i];
				//dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
			}
		}
		printf("%d\n",dp[m]);
	}
}