HDU-2602-Bone Collector （最基础DP！！）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31922    Accepted Submission(s): 13138

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

 

相信大家都对DP很熟了...但是我不是很熟.....上次北京现场赛就考了好多DP....差点压制死:-(

先贴个代码纪念纪念：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int dp[1005][1005], val[1005], vol[1005];

int main()
{
	int n, v, T;
	scanf("%d", &T);
	while(T--)
	{
		int i, j;
		memset(dp, 0, sizeof(dp));
		scanf("%d %d", &n, &v);
		for(i=1; i<=n; i++)
		{
			scanf("%d", &val[i]);
		}
		for(i=1; i<=n; i++)
		{
			scanf("%d", &vol[i]);
		}
		for(i=1; i<=n; i++)
		{
			for(j=0; j<=v; j++)
			{
				if(j >= vol[i]) dp[i][j] = max(dp[i-1][j], dp[i-1][j-vol[i]]+val[i]);
				else dp[i][j] = dp[i-1][j];
			}
		}
		printf("%d\n", dp[n][v]);
	}
	return 0;
}