【数据结构】bzoj1651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

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.

===================================华丽丽的分割线============================================

只要写一个支持区间修改和全局最大值查询的东西就好辣～

那不如直接写一个线段数暖手手～～～

这题好像可以直接差分然后就完了吧。。。

时间复杂度O(nlogn)，代码如下：

 #include <bits/stdc++.h>

 #define Maxn 1000007

 using namespace std;

 int read()

 {

     int x=,f=;char ch=getchar();

     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}

     while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

     return x*f;

 }

 struct seg

 {

     int lx,rx,mx,tag;

 };

 seg tree[Maxn*];

 int n;

 void build(int node, int l, int r)

 {

     tree[node].lx=l,tree[node].rx=r;

     tree[node].tag=,tree[node].mx=;

     if (l==r) return;

     int mid=(l+r)/;

     build(node*,l,mid);

     build(node*+,mid+,r);

 }

 void pushdown(int node)

 {

     if (tree[node].tag==) return;

     tree[node*].tag+=tree[node].tag;

     tree[node*].mx+=tree[node].tag;

     tree[node*+].tag+=tree[node].tag;

     tree[node*+].mx+=tree[node].tag;

     tree[node].tag=;

 }

 void update(int node, int l, int r, int del)

 {

     if (tree[node].rx<l) return;

     if (tree[node].lx>r) return;

     if (tree[node].lx>=l&&tree[node].rx<=r)

     {

         tree[node].tag+=del;

         tree[node].mx+=del;

         return;

     }

     pushdown(node);

     update(node*,l,r,del);

     update(node*+,l,r,del);

     tree[node].mx=max(tree[node*].mx,tree[node*+].mx);

 }

 int main()

 {

     n=read();

     build(,,);

     for (int i=;i<=n;i++)

     {

         int x=read(),y=read();

         update(,x,y,);

     }

     printf("%d\n",tree[].mx);

     return ;

 }