POJ 3624 Charm Bracelet(01背包裸题)
2017-05-08 22:50
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Charm Bracelet
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 38909 | Accepted: 16862 |
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
USACO 2007 December Silver
题目链接:http://poj.org/problem?id=3624
分析:01背包裸题,顺带敲了一遍,做个复习吧,怕忘了!详解请参看我的博客http://www.cnblogs.com/ECJTUACM-873284962/p/6815610.html,里面有对01背包的完全介绍以及用法!
下面给出AC代码:
#include <iostream> #include <algorithm> #include <stdio.h> using namespace std; int w[35000],d[35000],dp[35000]; int main() { int n,m; while(scanf("%d%d",&n,&m)!=EOF) { 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--) { dp[j]=max(dp[j],dp[j-w[i]]+d[i]); } } printf("%d\n",dp[m]); } return 0; }
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