POJ-1456 Supermarket（贪心，并查集优化）

Supermarket Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10725 Accepted: 4688 Description

A supermarket has a set Prod of products on sale. It earns a profit px for each product x∈Prod sold by a deadline dx that is measured as an integral number of time units starting from the moment the sale begins. Each product takes precisely one unit of time for being sold. A selling schedule is an ordered subset of products Sell ≤ Prod such that the selling of each product x∈Sell, according to the ordering of Sell, completes before the deadline dx or just when dx expires. The profit of the selling schedule is Profit(Sell)=Σx∈Sellpx. An optimal selling schedule is a schedule with a maximum profit. For example, consider the products Prod={a,b,c,d} with (pa,da)=(50,2), (pb,db)=(10,1), (pc,dc)=(20,2), and (pd,dd)=(30,1). The possible selling schedules are listed in table 1. For instance, the schedule Sell={d,a} shows that the selling of product d starts at time 0 and ends at time 1, while the selling of product a starts at time 1 and ends at time 2. Each of these products is sold by its deadline. Sell is the optimal schedule and its profit is 80.

Write a program that reads sets of products from an input text file and computes the profit of an optimal selling schedule for each set of products. Input

A set of products starts with an integer 0 <= n <= 10000, which is the number of products in the set, and continues with n pairs pi di of integers, 1 <= pi <= 10000 and 1 <= di <= 10000, that designate the profit and the selling deadline of the i-th product. White spaces can occur freely in input. Input data terminate with an end of file and are guaranteed correct. Output

For each set of products, the program prints on the standard output the profit of an optimal selling schedule for the set. Each result is printed from the beginning of a separate line. Sample Input

4 50 2 10 1 20 2 30 1

7 20 1 2 1 10 3 100 2 8 2 5 20 50 10 Sample Output

80 185

题意： 超市有n个商品，每个商品都是利润和保质期，每天只卖一种，问，怎么卖才可以获得最大利润，求出最大利润 思路： 贪心，这个题目我一开始看到，觉得可以用背包做啊，和背包题目很像啊。为什么这道题目可以用贪心做呢？这里说一下我对贪心和动态规划的理解：贪心就是每次都选取最优的结果就会是最优的，比如遇到背包问题，如果每个物品的体积都是1，那么就完全可以用贪心解决。但是如果每个物品的体积不一样呢？就不可以用贪心，贪心会错，自己体会。这是动态规划和贪心的区别。这道题目可以用贪心做，这里要贪两个地方，首先我先卖利润最大的商品，这是第一个贪。其次利润最大的商品应该尽量在后面卖，接近保质期的时间卖，这样前面的时间可以多卖一些其他的商品让利润最大，这是第二个贪。这道题目由于每个商品卖的时间都是1天，如果时间不固定，就不可以用贪心。和前面的道理是一样的，这里给一个另一个hdu的题目的博客，就可以体会到贪心和动态规划的区别。 http://blog.csdn.net/Dacc123/article/details/50397213 这道题目还有一个用并查集优化的方法，算是对并查集的一个应用。 贪心：

#include <iostream>
#include <math.h>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <stdlib.h>

using namespace std;
#define  MAX 100000
int n;
int vis[MAX+5];
struct Node
{
    int price;
    int time;
}a[MAX+5];
int cmp(Node a,Node b)
{
    if(a.price==b.price)
        return a.time<b.time;
    return a.price>b.price;
}
int main()
{
    int ans;
    while(scanf("%d",&n)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        ans=0;
        for(int i=1;i<=n;i++)
            scanf("%d%d",&a[i].price,&a[i].time);
        sort(a+1,a+n+1,cmp);
        for(int i=1;i<=n;i++)
        {
           for(int j=a[i].time;j>=1;j--)
           {
               if(!vis[j])
               {
                   ans+=a[i].price;
                   vis[j]=1;
                   break;
               }
           }
        }
        printf("%d\n",ans);
    }
}

并查集优化：

#include <iostream>
#include <math.h>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <stdlib.h>

using namespace std;
#define  MAX 100000
int n;
int vis[MAX+5];
struct Node
{
    int price;
    int time;
}a[MAX+5];
int cmp(Node a,Node b)
{
  return a.price>b.price;
}
int father[MAX+5];
int find(int x)
{
    if(x!=father[x])
        father[x]=find(father[x]);
    return father[x];
}
void init()
{
    for(int i=0;i<=MAX;i++)
        father[i]=i;
}
int main()
{
    int ans;
    while(scanf("%d",&n)!=EOF)
    {
        init();
        ans=0;
        for(int i=1;i<=n;i++)
            scanf("%d%d",&a[i].price,&a[i].time);
        sort(a+1,a+n+1,cmp);
        for(int i=1;i<=n;i++)
        {
             int b;
             b=find(a[i].time);
             if(b>0)
             {
                 ans+=a[i].price;
                 father[b]=b-1;
             }
        }
        printf("%d\n",ans);
    }
}