poj3624

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19451		Accepted: 8842

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 W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 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: W_i and D_i

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

很简单的一道01背包，但是应该注意，不要多次输入，我就被坑了

#include<iostream>
#include<stdio.h>
#include<string.h>
int va[12885],w[12885],dp[12885];
using namespace std;
int main()
{
    int i,j,n,v;
    scanf("%d%d",&n,&v);
        memset(dp,0,sizeof(dp));
        for(i=0; i<n; i++)
            scanf("%d%d",&w[i],&va[i]);
        for(i=0; i<n; i++)
            for(j=v; j>=w[i]; j--)
                dp[j]=max(dp[j],dp[j-w[i]]+va[i]);
        printf("%d\n",dp[v]);
    return 0;
}