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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 6 1 4 2 6 3 12 2 7
Sample Output
23
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 using namespace std; 5 6 int main() 7 { 8 int w[3500]; 9 int c[3500]; 10 int dp[13000]; 11 int n,V; 12 while(scanf("%d%d",&n,&V)!=EOF) 13 { 14 memset(dp,0,sizeof(dp)); 15 for(int i=0;i<n;i++) 16 { 17 scanf("%d",&c[i]); 18 scanf("%d",&w[i]); 19 } 20 for(int i=0;i<n;i++) 21 for(int v=V;v>=0;v--) 22 if(v-c[i]>=0) 23 dp[v]=max(dp[v],dp[v-c[i]]+w[i]); 24 printf("%d\n",dp[V]); 25 } 26 return 0; 27 }
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原文地址：http://www.cnblogs.com/homura/p/4764316.html
