POJ 3624 Charm Bracelet(01背包)

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34532		Accepted: 15301

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

Sample Output

Source

USACO 2007 December Silver

感觉数据有点水，典型01背包

 1 //Accepted    212K    266MS    C++    395B
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<string.h>
 5 using namespace std;
 6 
 7 int dp[13000];
 8 int w[3500], d[3500];
 9 int main(void)
10 {
11     int n,m;
12     while(scanf("%d%d", &n,&m)!=EOF){
13         for(int i=0;i<n;i++)
14             scanf("%d%d",&w[i],&d[i]);
15         memset(dp, 0 ,sizeof(dp));
16         for(int i=0;i<n;i++)
17             for(int j=m;j>=w[i];j--)
18                 dp[j] = max(dp[j], dp[j-w[i]] + d[i]);
19         printf("%d
", dp[m]);
20     }
21     return 0;
22 }