POJ 3624 Charm Bracelet

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34193		Accepted: 15154

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

 

  01背包问题。

 

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

const int MAX_N=3500;
const int MAX_M=13000;

int v[MAX_N],w[MAX_N];
int f[MAX_M];
int main()
{
    int N,M,V,W;
    while(scanf("%d%d",&N,&M)==2){
        memset(f,0,sizeof(f));
        for(int i=1;i<=N;i++){
            scanf("%d%d",&V,&W);
            for(int j=M;j>=0;j--)
                if(j>=V) f[j]=max(f[j], f[j-V]+W);
        }
        printf("%d
",f[M]);
    }
}