【贪心】POJ3190:Stall Reservations

Description

Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviously, FJ must create a reservation system to determine which stall each cow can be assigned for her milking time. Of course, no cow will share such a private moment with other cows. 

Help FJ by determining:

• The minimum number of stalls required in the barn so that each cow can have her private milking period
• An assignment of cows to these stalls over time

Many answers are correct for each test dataset; a program will grade your answer.

Input

Line 1: A single integer, N 

Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.

Output

Line 1: The minimum number of stalls the barn must have. 

Lines 2..N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.

Sample Input

5
1 10
2 4
3 6
5 8
4 7

Sample Output

4
1
2
3
2
4

Hint

Explanation of the sample: 

Here's a graphical schedule for this output: 

Time     1  2  3  4  5  6  7  8  9 10

Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>

Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..

Stall 3 .. .. c3>>>>>>>>> .. .. .. ..

Stall 4 .. .. .. c5>>>>>>>>> .. .. ..

Other outputs using the same number of stalls are possible.

 

大致题意：

有一群牛，每个牛都有固定的吃饭时间，并且一个牛槽只能同时让一只牛吃饭。给出所有牛吃饭的开始时间和结束时间，求出最优的策略，使用最少的牛槽让所有的牛吃上饭。输出最少的牛槽数，以及每个牛在第几个牛槽吃。

大体思路：

贪心。用到了优先队列。

1)将牛按照开始时间由小到大排序，开始时间相等的话按结束时间由小到大排序。

2)创建一个优先队列，在队列中牛按照结束的时间从小到大排序。

3)把第一只牛加入队列，然后检查第二只牛能否和它同用一槽。如果能，第一只牛出队列，第二只牛进队列。若不能，第二只牛进队列，牛槽数加一。继续该步骤，知道所有的牛都吃过了。

4）注意，记录牛的数据时，除了开始和结束时间，还要记录牛一开始的顺序，以便记录它在哪个槽吃。

代码如下：

#include <iostream>
#include <queue>
#include <algorithm>
#include <cstdio>
using namespace std;
const int maxn = 50010;
struct node
{
    int st,en,pos;
    friend const bool operator < (const node a ,const node b)
    {
        if(a.en==b.en)
            return a.st>b.st;
        return a.en>b.en;
    }
};
node cow[maxn];
bool cmp(node a,node b)
{
    if(a.st==b.st)
        return a.en<b.en;
    return a.st<b.st;
}
priority_queue<node> q;
int ans[maxn];
int n;
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&cow[i].st,&cow[i].en);
            cow[i].pos=i;
        }
        sort(cow+1,cow+1+n,cmp);
        q.push(cow[1]);
        int cnt=1;
        ans[cow[1].pos]=cnt;
        for(int i=2;i<=n;i++)
        {
            if(q.top().en<cow[i].st)
            {
                ans[cow[i].pos]=ans[q.top().pos];
                q.pop();
            }
            else
            {
                cnt++;
                ans[cow[i].pos]=cnt;
            }
            q.push(cow[i]);
        }
        printf("%d
",cnt);
        for(int i=1;i<=n;i++)
        {
            printf("%d
",ans[i]);
        }
    }
    return 0;
}