| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 19451 | Accepted: 8842 |
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
很简单的一道01背包，但是应该注意，不要多次输入，我就被坑了
1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 int va[12885],w[12885],dp[12885]; 5 using namespace std; 6 int main() 7 { 8 int i,j,n,v; 9 scanf("%d%d",&n,&v); 10 memset(dp,0,sizeof(dp)); 11 for(i=0; i<n; i++) 12 scanf("%d%d",&w[i],&va[i]); 13 for(i=0; i<n; i++) 14 for(j=v; j>=w[i]; j--) 15 dp[j]=max(dp[j],dp[j-w[i]]+va[i]); 16 printf("%d ",dp[v]); 17 return 0; 18 }