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

二次联通门 : BZOJ 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

 

 

 

权限题放题面

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

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

    差分的思想真的是妙啊
    通过给两个点打标记,实现给整个区间打标记
    
        可惜之前一直不知道这个东西
        那么有些题不就变得很水了吗。。。    
*/
#include <cstdio>
#include <iostream>

const int BUF = 12312313;
char Buf[BUF], *buf = Buf;

inline void read (int &now)
{
    for (now = 0; !isdigit (*buf); ++ buf);
    for (; isdigit (*buf); now = now * 10 + *buf - '0', ++ buf);
}
#define Max 1000200
int key[Max];
inline int max (int a, int b)
{
    return a > b ? a : b;
}
int Main ()
{
    fread (buf, 1, BUF, stdin);
    int N, x, y; read (N); register int i;
    for (i = 1; i <= N; ++ i)
    {
        read (x), read (y);
        ++ key[x], -- key[y + 1];
    }
    int Answer = -1, res = 0;
    for (i = 1; i <= Max; ++ i)
    {
        res += key[i];
        Answer = max (Answer, res);
    }
    printf ("%d", Answer);

    return 0;
}
int ZlycerQan = Main ();
int main (int argc, char *argv[]) {;}

 

 

 

posted @ 2017-08-22 08:24  ZlycerQan  阅读(198)  评论(0编辑  收藏  举报