标签：arc mis 博客 nts desc 代码 describes char gone
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 38909 | Accepted: 16862 |
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
1 #include <iostream> 2 #include <algorithm> 3 #include <stdio.h> 4 using namespace std; 5 int w[35000],d[35000],dp[35000]; 6 int main() 7 { 8 int n,m; 9 while(scanf("%d%d",&n,&m)!=EOF) 10 { 11 for(int i=1;i<=n;i++) 12 scanf("%d%d",&w[i],&d[i]); 13 for(int i=1;i<=n;i++) 14 { 15 for(int j=m;j>=w[i];j--) 16 { 17 dp[j]=max(dp[j],dp[j-w[i]]+d[i]); 18 } 19 } 20 printf("%d\n",dp[m]); 21 } 22 return 0; 23 }
POJ 3624 Charm Bracelet(01背包裸题)
标签：arc mis 博客 nts desc 代码 describes char gone
原文地址：http://www.cnblogs.com/ECJTUACM-873284962/p/6828362.html
