POJ 3624 Charm Bracelet【01背包入门题】

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分类专栏： POJ ACM_DP ACM_HDU刷题录文章标签： POJ3624 Charm Bracelet 01背包入门题

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Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34333		Accepted: 15215

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

Sample Output

Source

USACO 2007 December Silver

原题链接：http://poj.org/problem?id=3624

01背包入门题，不解释。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=12880+5;
int v[maxn];
int w[maxn];
int dp[maxn];
int main()
{
    int n,m;
    ios::sync_with_stdio(false);
    cin.tie(0);
    //freopen("data/3624.txt","r",stdin);
    while(cin>>n>>m)
    {
        for(int i=0; i<n; i++)
            cin>>w[i]>>v[i];
        memset(dp,0,sizeof(dp));

        for(int i=0; i<n; i++)
        {
            for(int j=m; j>=w[i]; j--)
            {
                dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
            }
        }
        cout<<dp[m]<<endl;

    }
    return 0;
}