POJ 3624 Charm Bracelet

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 41212 Accepted: 17929

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

01背包水题

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[40010];int D[40010];int W[40010];int main(){//freopen("in.txt","r",stdin);int n,m;while(cin>>n>>m){memset(dp,0,sizeof(dp));for(int i=1;i<=n;i++) scanf("%d%d",&W[i],&D[i]);for(int i=1;i<=n;i++){for(int j=m;j>=W[i];j--){dp[j]=max(dp[j],dp[j-W[i]]+D[i]);}}//cout<<dp[m]<<endl;int ans=-1;for(int i=1;i<=m;i++) ans=max(ans,dp[i]);cout<<ans<<endl;}}