Hduoj2602【01背包】

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31842    Accepted Submission(s): 13104

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
   
   
   
   

 

Author

Teddy

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest 

#include<stdio.h>
#include<string.h>
#define  max(A,B)  (A)>(B)?A:B
int main()
{
	int T;
	while(scanf("%d", &T) != EOF){
	while(T--)
	{
		int n, V, vo[1010], va[1010];
		scanf("%d%d", &n, &V);
		for(int i = 1; i <= n; i++)
		scanf("%d", &va[i]);
		for(int i = 1; i <= n; i++)
		scanf("%d", &vo[i]);
		int f[1010];
		memset(f, 0, sizeof(f));
		for(int i = 1; i <= n; i++)
		{
			for(int j = V; j >= vo[i]; j--)
			{
				f[j] = max(f[j], f[j - vo[i]] + va[i]);
			}
		} 
		printf("%d\n", f[V]); 
	}
}
	return 0;
} 

题意：给出一个容量为V的背包，用来收集骨头，每个骨头占据一定的容量，并且每一个骨头都有一个价值，现给出一系列骨头。求怎么收集能得到最大的价值。

思路：标准的01背包问题。