Bone Collector (HDU_2602) 01背包

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 44702    Accepted Submission(s): 18620

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 2 ³¹).

 

Sample Input

   
   
   
   

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

 

Sample Output

   
   
   
   

    
    
    
    14
   
   
   
   

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602

题目大意：有n个不同骨头，每块占用一定空间以及具有一定的价值，给定一个容量，求最多能装进的骨头的价值是多少.

解题思路：01背包。

    动态转移方程：dp[i][j]=max(dp[i][j],dp[i+1][j-volume[i]]+value[i]);

  d[i][j]代表在容量为j的情况下，不取i所能获得的最大价值。

   dp[i+1][j-volume[i]]+value[i]代表在容量为j的情况下，取i所获得的最大价值。

#include"iostream"
#include"cstdio"
using namespace std;
int value[1005];
int volume[1005];
int dp[1005][1005];
int max(int a,int b)
{
	return a>b?a:b;
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--){
		int n,v;
		scanf("%d%d",&n,&v);
		for(int i=0;i<=n+1;i++)
			for(int j=0;j<=v;j++)
				dp[i][j]=0;
		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=n;i>=1;i--){
			for(int j=0;j<=v;j++){
				dp[i][j]=(i==n?0:dp[i+1][j]);
				if(j>=volume[i])
					dp[i][j]=max(dp[i][j],dp[i+1][j-volume[i]]+value[i]);
			}
		}
		printf("%d\n",dp[1][v]);
	}
	return 0;
 }