POJ——3624 Charm Bracelet

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 27943		Accepted: 12589

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 weightW_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

0 1背包

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<map>
#include<numeric>
#include<stack>
#include<list>
const int INF=1<<30;
const int inf=-(1<<30);
const int MAX=100010;


using namespace std;


struct data
{
    int x,y;
} a[10010];
int dp[100010];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(dp,0,sizeof(dp));
        for(int i=1; i <= n; i++)
            scanf("%d%d",&a[i].x,&a[i].y);
        for(int i=1;i <= n; i++)
            for(int j=m; j>=a[i].x; j--)
                dp[j]=max(dp[j],dp[j-a[i].x]+a[i].y);
        cout<<dp[m]<<endl;
    }
}