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POJ3624Charm Bracelet（01背包）
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 23359 | Accepted: 10532 |
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 theN (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weightWi (1 ≤ Wi ≤ 400), a ‘desirability‘ factorDi (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM (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
还是01背包
#include<iostream>#include<cstdio>#include<cmath>#include<queue>#include<algorithm>#include<cstring>using namespace std;int main(){ int n,i,j,v; while(scanf("%d%d",&n,&v)!=EOF){ int dp[13000]={0},c[3500],w[3500]; for(i=0;i<n;i++) scanf("%d%d",c+i,w+i); for(i=0;i<n;i++) { for(j=v;j>=c[i];j--) { if(dp[j]<dp[j-c[i]]+w[i]) { dp[j]=dp[j-c[i]]+w[i]; } } } printf("%d\n",dp[v]); } return 0;}