poj3624——Charm Bracelet

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

非常基础的一道01背包

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#define _clr(x,a) memset(x,a,sizeof(x));
using namespace std;
const int N=15000;
int v[N];
int w[N];
int dp[N];
int main(void){
    //freopen("data.txt","r",stdin);
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        _clr(v,0);
        _clr(w,0);
        _clr(dp,0);
        for(int i=0;i<n;i++){
            scanf("%d%d",&v[i],&w[i]);
        }
        for(int i=0;i<n;i++){
            for(int j=m;j>=v[i];j--){
                dp[j]=max(dp[j],dp[j-v[i]]+w[i]);
            }
        }
        printf("%d\n",dp[m]);
    }
    return 0;
}