Charm Bracelet (POJ 3624）

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 40141		Accepted: 17439

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

用二维动态肯定不行，这就需要优化成一维的了。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int N = 3410;
int w[N],d[N];
int dp[12881];
int main(){
    int n,m;
    cin>>n>>m;
    for(int i = 0; i < n; i ++)scanf("%d %d",&w[i],&d[i]);
    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;
}