Stall Reservations 专用牛棚

Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over s
ome precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviou
sly, FJ must create a reservation system to determine which stall each cow can be assigned for her m
ilking time. Of course, no cow will share such a private moment with other cows. Help FJ by determin
ing: * The minimum number of stalls required in the barn so that each cow can have her private milki
ng period * An assignment of cows to these stalls over time

有N头牛,每头牛有个喝水时间,这段时间它将专用一个Stall 现在给出每头牛的喝水时间段,问至少要多少个Stall
才能满足它们的要求

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
//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.

sol:差分思想，简单题，直接看标程吧。

#include<cstdio>
using namespace std;
int a[1000010];
int main()
{
    int ans=-1;
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        a[x]+=1;a[y+1]-=1;
    }
    int length=0;
    for(int i=1;i<=1000000;i++)
    {
        length+=a[i];
        if(length>ans)ans=length;
    }
    printf("%d",ans);
    return 0;
}