标签：ble plm span center desc ack const cst iostream
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 41106 | Accepted: 17884 |
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 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
Source
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n,m,w[3500],d[3500],f[13000]; int main(){ scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) scanf("%d%d",&w[i],&d[i]); for(int i=1;i<=n;i++) for(int j=m;j>=w[i];j--) f[j]=max(f[j],f[j-w[i]]+d[i]); printf("%d",f[m]); }
标签：ble plm span center desc ack const cst iostream
原文地址：http://www.cnblogs.com/cangT-Tlan/p/7434464.html
