POJ 3624 背包水题
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POJ 3624
01背包水题。
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 constraint
Sample Input
- 4 6
- 1 4
- 2 6
- 3 12
- 2 7
Sample Output
- 23
代码
#include<iostream>#include<cstdio>#include<cmath>using namespace std;int dp[12900];int main(){ int n,m; scanf("%d%d",&n,&m); sizeof(dp,0,sizeof(dp)); int w[3500]; int d[3500]; for (int i = 0;i<n;i++) scanf("%d%d",&w[i],&d[i]); for (int i = 0;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]);}
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