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

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题目

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1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

Time Limit: 10 Sec  Memory Limit: 64 MB
Submit: 553  Solved: 307
[Submit][Status]

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

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题解

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这道题用时间戳的思路就可以了，我们统计同一时间最大的时间戳个数就是答案。

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代码

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 /*Author:WNJXYK*/
 #include<cstdio>
 using namespace std;

 #define LL long long
 #define Inf 2147483647
 #define InfL 10000000000LL

 inline int abs(int x){if (x<) return -x;return x;}
 inline int abs(LL x){if (x<) return -x;return x;}
 inline void swap(int &x,int &y){int tmp=x;x=y;y=tmp;}
 inline void swap(LL &x,LL &y){LL tmp=x;x=y;y=tmp;}
 inline int remin(int a,int b){if (a<b) return a;return b;}
 inline int remax(int a,int b){if (a>b) return a;return b;}
 inline LL remin(LL a,LL b){if (a<b) return a;return b;}
 inline LL remax(LL a,LL b){if (a>b) return a;return b;}
 inline void read(int &x){x=;int f=;char ch=getchar();while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}x=x*f;}
 inline void read(LL &x){x=;LL f=;char ch=getchar();while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}x=x*f;}
 inline void read(int &x,int &y){read(x);read(y);}
 inline void read(LL &x,LL &y){read(x);read(y);}
 inline void read(int &x,int &y,int &z){read(x,y);read(z);}
 inline void read(int &x,int &y,int &n,int &m){read(x,y);read(n,m);}
 inline void read(LL &x,LL &y,LL &z){read(x,y);read(z);}
 inline void read(LL &x,LL &y,LL &n,LL &m){read(x,y);read(n,m);}
 const int Maxn=;
 int n;
 int a,b;
 int t[Maxn+];
 int main(){
     read(n);
     for (;n--;){
         read(a,b);
         t[a]++;
         t[b+]--;
     }
     int Ans=;
     for (int i=;i<=Maxn;i++){
         t[i]=t[i-]+t[i];
         Ans=remax(Ans,t[i]);
     }
     printf("%d\n",Ans);
     return ;
 }