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最新推荐文章于 2016-12-11 16:30:00 发布

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Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11298		Accepted: 5130

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

这个题目是一个典型的0-1背包问题。。

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int N,M;
int cost[4000];
int weight[4000];
int f[4000*400];
void init(int n)
{
    for(int i=1;i<=n;i++)
    {
        cin>>cost[i]>>weight[i];
    }
}
void zeroonepack(int c,int w)
{
    for(int v=M;v>=c;v--)
    {
        f[v]=max(f[v],f[v-c]+w);
    }
}
int main()
{
    cin>>N>>M;
    init(N);
    memset(f,0,sizeof(f));
    for(int i=1;i<=N;i++)
    {
        zeroonepack(cost[i],weight[i]);
    }
    cout<<f[M]<<endl;
    return 0;
}