Charm Bracelet

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 48555   Accepted: 20584

Description

Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di

Output

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

Sample Input

4 6
1 4
2 6
3 12
2 7

Sample Output

23

思路：

     dp模板题了吧。状态转移方程：dp[j] = maxn( dp[j] ,dp[j-w[i]] + v[i])。

代码：

#include<stdio.h>
#include<string.h>
int maxn(int a,int b)
{
	if(a>=b)
		return a;
	return b;
}
int main()
{
	int i,j,n,m;
	int w[3500],v[3500];
	int dp[13000];
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(dp,0,sizeof(dp));
		for(i=1;i<=n;i++)
			scanf("%d%d",&w[i],&v[i]);
		for(i=1;i<=n;i++)
		{
			for(j=m;j>=w[i];j--)
			{
				dp[j]=maxn(dp[j],dp[j-w[i]]+v[i]);
			}
		}
		printf("%d\n",dp[m]);
	}
	return 0;
}