poj 3190 Stall Reservations

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

Sample Output

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.

贪心思想，先把输入的时间段升序排序，创建一个优先队列，优先知道最靠左时间段，方便共用stall。

代码：

#include <iostream>
#include <algorithm>
#include <queue>
#include <cstdio>
using namespace std;

class time
{
    public:
    int a,b;
    int id;
    friend bool operator <(time x,time y)
    {
        if(x.b==y.b)return x.a>y.a;///右端点相同 则取左端点靠前的 即最先被分配的
        return x.b>y.b;///堆顶元素，右端点最小
    }
}s[50001];
int cmp(time x,time y)
{
    if(x.a==y.a)return x.b < y.b;///左端点相同的 右端点也升序
    return x.a < y.a;///左端点升序排序
}
int main()
{
    int n,used[50001],ans = 0;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&s[i].a,&s[i].b);
        s[i].id=i;
    }
    sort(s,s+n,cmp);
    priority_queue<time> q;

    for(int i = 0;i < n;i ++)
    {
        if(!q.empty()&&s[i].a > q.top().b)///和队顶比较  队顶保存的是最靠前的，如果满足条件可以插在它的后面，公用一个stall
        {
            used[s[i].id]=used[q.top().id];
            q.pop();
        }
        else///否则另加一个stall
        {
            ans++;
            used[s[i].id]=ans;
        }
        q.push(s[i]);
    }
    cout<<ans<<endl;
    for(int i=0;i<n;i++)
        cout<<used[i]<<endl;
}

View Code