POJ3624解题报告

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7247 Accepted: 3278

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

题意：给定每个物品的重量和价值，求给定最大重量下能获得的最大价值。。

我的背包第一题

思路：01背包，状态方程为dp[i][j]=MAX(dp[i-1][j],dp[i-1][j-weight[i]]+value[i])

#include<iostream>using namespace std;int N,W,weight[40000],cost[40000],dp[40000],i,j;int main(){while(cin>>N>>W&&N){for(i=0;i<N;i++)cin>>cost[i]>>weight[i];memset(dp,0,sizeof(dp));for(i=0;i<N;i++)for(j=W;j>=cost[i];j--){if(j-cost[i]>=0){if(dp[j-cost[i]]+weight[i]>dp[j])dp[j]=dp[j-cost[i]]+weight[i];}}cout<<dp[W]<<endl;}}