标签：
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.
* 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
* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
4 6 1 4 2 6 3 12 2 7
23
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 int dp[12900],w[3500],d[3500],n,m; 6 int main(){ 7 while(~scanf("%d %d",&n,&m)){ 8 for(int i = 0; i < n; ++i) 9 scanf("%d %d",w+i,d+i); 10 memset(dp,0,sizeof dp); 11 for(int i = 0; i < n; ++i) 12 for(int j = m; j >= w[i]; --j) 13 dp[j] = max(dp[j],dp[j-w[i]] + d[i]); 14 printf("%d\n",dp[m]); 15 } 16 return 0; 17 }
标签：
原文地址：http://www.cnblogs.com/crackpotisback/p/4415715.html
