POJ 3624 Charm Bracelet(背包)
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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 61 42 63 122 7
Sample Output
23
Source
裸的背包
代码:
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=30000+1000;int dp[maxn];int main(){ int n,m,w,v; while(~scanf("%d%d",&n,&m)) { memset(dp,0,sizeof(dp)); for(int i=0;i<n;i++) { scanf("%d%d",&w,&v); for(int j=m;j>=w;j--) { dp[j]=max(dp[j-w]+v,dp[j]); } } int ans=0; for(int i=0;i<=m;i++) ans=max(ans,dp[i]); printf("%d\n",ans); } return 0;}
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