POJ3190-Stall Reservations（区间贪心+优先队列）

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13353		Accepted: 4685		Special Judge

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.

Source

USACO 2006 February Silver

分析：

用优先队列，如果说这个时候能够空出畜栏，就继承之前这个畜栏的编号，

如果说不能空出畜栏就增加畜栏的数目，畜栏的编号是i

代码：

#include
#include
#include
#include
#include
#include
using namespace std;
struct Cow
{
    int a,b;
    int No;
    bool operator <(const Cow &c) const
    {
        return as.end;
    }
    stall (int e,int n):end(e),No(n){ }
    };
int main()
{
    int n;
    scanf("%d",&n);
    for(int  i = 0 ;i pq;
    for(int  i =0;i