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

Time Limit: 10 Sec  Memory Limit: 64 MB
Submit: 638  Solved: 359
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## 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.

## Source

Silver

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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<deque>
using namespace std;
int a[1000005];
int main(){
int n;
int nmax=-1;
scanf("%d",&n);
for(int i=1,o,e;i<=n;i++){
scanf("%d%d",&o,&e);
a[o]++;a[e+1]--;
nmax=max(nmax,e);
}
int tmp=0;int ans=-1;
for(int i=1;i<=nmax;i++){
tmp+=a[i];
if(tmp>ans)
ans=tmp;
}
printf("%d\n",ans);
return 0;
}

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