POJ 3624.Charm Bracelet-动态规划0-1背包

Charm Bracelet
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 47876   Accepted: 20346

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

USACO 2007 December Silver
 
 
 
代码:
 1 #include<cstdio>
 2 using namespace std;
 3 const int N=4000;
 4 int maxx(int x,int y){
 5     return x>y?x:y;
 6 }
 7 
 8 int main(){
 9     int n,m,i,j;
10     int w[N],p[N];
11     while(~scanf("%d%d",&n,&m)){
12         int c[20000]={0};
13         for(int i=1;i<=n;i++)
14             scanf("%d%d",&w[i],&p[i]);
15         for(int i=1;i<=n;i++)
16             for(int j=m;j>=w[i];j--)
17         c[j]=maxx(c[j],c[j-w[i]]+p[i]);
18         printf("%d\n",c[m]);
19     }
20      return 0;
21 }

 

 

 

 

posted @ 2017-07-22 21:32  ZERO-  阅读(178)  评论(0)  编辑  收藏  举报