POJ 3624.Charm Bracelet-动态规划0-1背包

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 47876		Accepted: 20346

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

 

 

 

代码:

#include<cstdio>
using namespace std;
const int N=4000;
int maxx(int x,int y){
    return x>y?x:y;
}

int main(){
    int n,m,i,j;
    int w[N],p[N];
    while(~scanf("%d%d",&n,&m)){
        int c[20000]={0};
        for(int i=1;i<=n;i++)
            scanf("%d%d",&w[i],&p[i]);
        for(int i=1;i<=n;i++)
            for(int j=m;j>=w[i];j--)
        c[j]=maxx(c[j],c[j-w[i]]+p[i]);
        printf("%d
",c[m]);
    }
     return 0;
}