# BZOJ 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

## 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

## 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.

5
1 10
2 4
3 6
5 8
4 7

## Sample Output

4

OUTPUT DETAILS:

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.

## HINT

/*
BZOJ 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

差分的思想真的是妙啊
通过给两个点打标记，实现给整个区间打标记

可惜之前一直不知道这个东西
那么有些题不就变得很水了吗。。。
*/
#include <cstdio>
#include <iostream>

const int BUF = 12312313;
char Buf[BUF], *buf = Buf;

{
for (now = 0; !isdigit (*buf); ++ buf);
for (; isdigit (*buf); now = now * 10 + *buf - '0', ++ buf);
}
#define Max 1000200
int key[Max];
inline int max (int a, int b)
{
return a > b ? a : b;
}
int Main ()
{
int N, x, y; read (N); register int i;
for (i = 1; i <= N; ++ i)
{
++ key[x], -- key[y + 1];
}
int Answer = -1, res = 0;
for (i = 1; i <= Max; ++ i)
{
res += key[i];
}
int main (int argc, char *argv[]) {;}