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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 34193 | Accepted: 15154 |
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
Source
1 #include <iostream> 2 #include <algorithm> 3 #include <cstring> 4 using namespace std; 5 6 const int MAX_N=3500; 7 const int MAX_M=13000; 8 9 int v[MAX_N],w[MAX_N]; 10 int f[MAX_M]; 11 int main() 12 { 13 int N,M,V,W; 14 while(scanf("%d%d",&N,&M)==2){ 15 memset(f,0,sizeof(f)); 16 for(int i=1;i<=N;i++){ 17 scanf("%d%d",&V,&W); 18 for(int j=M;j>=0;j--) 19 if(j>=V) f[j]=max(f[j], f[j-V]+W); 20 } 21 printf("%d\n",f[M]); 22 } 23 }
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原文地址：http://www.cnblogs.com/cumulonimbus/p/5830659.html
