程序師世界是廣大編程愛好者互助、分享、學習的平台,程序師世界有你更精彩!
首頁
編程語言
C語言|JAVA編程
Python編程
網頁編程
ASP編程|PHP編程
JSP編程
數據庫知識
MYSQL數據庫|SqlServer數據庫
Oracle數據庫|DB2數據庫
 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> HDU2602 Bone Collector

HDU2602 Bone Collector

編輯:C++入門知識


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
1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output
14

#include 
using namespace std;

int volume[1005];
int value[1005];
int dp[1005];
int n,m;

int main()
{
	freopen("C:\\in.txt","r",stdin);
	int T;
	scanf("%d",&T);
	while(T--){
		int N,V;
		memset(dp,0,sizeof(dp));
		scanf("%d %d",&N,&V);
		for(int i=1;i<=N;i++)
			scanf("%d",&value[i]);
		for(int i=1;i<=N;i++)
			scanf("%d",&volume[i]);
		for(int i=1;i<=N;i++)
			for(int j=V;j>=volume[i];j--)
				dp[j]=max(dp[j],dp[j-volume[i]]+value[i]);
		printf("%d\n",dp[V]);
	}
    return 0;
}



  1. 上一頁:
  2. 下一頁:
Copyright © 程式師世界 All Rights Reserved