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hdu 2602 - Bone Collector(01背包)解题报告

时间:2015-08-10 22:08:24      阅读:116      评论:0      收藏:0      [点我收藏+]

标签:01背包

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 39532    Accepted Submission(s): 16385


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 ?
技术分享

 

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
1 5 10 1 2 3 4 5 5 4 3 2 1
 

Sample Output
14
 

题解:

    裸的01背包,经典例题。

参考代码:

#include<stdio.h>
#define M 1111
#include<string.h>
int max(int a,int b)
{
	return a>b?a:b;
}
int main()
{
	int t,a[M],b[M],dp[M];
	scanf("%d",&t);
	while(t--)
	{
		int n,v;
		memset(dp,0,sizeof(dp));
		scanf("%d%d",&n,&v);
		for(int i=0;i<n;i++)
		    scanf("%d",&b[i]);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<n;i++)
        {
        	for(int j=v;j>=a[i];j--)
        	{
	        	dp[j]=max(dp[j],dp[j-a[i]]+b[i]);
	        }
        }
        printf("%d\n",dp[v]);
	}
	return 0;
}



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hdu 2602 - Bone Collector(01背包)解题报告

标签:01背包

原文地址:http://blog.csdn.net/luwhere/article/details/47405567

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