bzoj1651[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

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


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

不妨试下这个数据,对于按结束点SORT,再GREEDY的做法 1 3 5 7 6 9 10 11 8 12 4 13 正确的输出应该是3

第一次写差分序列……挺好用的。

差分序列是一个类似前缀和的逆运算。就是对于数据a[i]，s[i]=a[i]-a[i-1]。则Σs[i]=a[i]。这样做的好处是，对于只要求区间加减一个数的操作，比如在区间[l,r]加上sum，那么只要在s[l]加上sum，在s[r+1]减去sum即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x*=10;x+=ch-'0';ch=getchar();}
    return x*f;
}
int a[1000010];
int main()
{
int n=read(),now=0,ans=0,mx=0;
for (int i=1;i<=n;i++)
{
int x,y;
cin>>x>>y;
a[x]++;a[y+1]--;
if (y>mx) mx=y;
}
for (int i=1;i<=mx;i++)
{
now+=a[i];
if (now>ans) ans=now;
}
printf("%d",ans);
}