POJ 3624 Charm Bracelet(01背包)

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34532		Accepted: 15301

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

Source

USACO 2007 December Silver

感觉数据有点水，典型01背包

//Accepted    212K    266MS    C++    395B
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

int dp[13000];
int w[3500], d[3500];
int main(void)
{
    int n,m;
    while(scanf("%d%d", &n,&m)!=EOF){
        for(int i=0;i<n;i++)
            scanf("%d%d",&w[i],&d[i]);
        memset(dp, 0 ,sizeof(dp));
        for(int i=0;i<n;i++)
            for(int j=m;j>=w[i];j--)
                dp[j] = max(dp[j], dp[j-w[i]] + d[i]);
        printf("%d
", dp[m]);
    }
    return 0;
}