POJ3624 Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 34471 Accepted: 15268

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 W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 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: W_i and D_i

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 61 42 63 122 7

Sample Output

23

背包裸题

#include<cstdio>  #include<cstring>  #include<algorithm>  using namespace std;  int dp[14040];  int val[4040],cost[4040];  int max(int x,int y){if(x>y)return x;elsereturn y;}int main()  {   int n,v;     while(scanf("%d%d",&n,&v)!=EOF )      {                  int i,j;          for(i=1;i<=n;i++) {    scanf("%d%d",&cost[i],&val[i]);    }         memset(dp,0,sizeof(dp));          for(i=1;i<=n;i++)          {              for(j=v;j>=cost[i];j--)              {                  dp[j]=max(dp[j],dp[j-cost[i]]+val[i]);                  }          }          printf("%d\n",dp[v]);      }  }