BZOJ 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚-优快云博客

本文介绍了解决BZOJ1651 StallReservations问题的方法，该问题是关于为特定时间段内需要使用独立隔间的奶牛分配最少隔间数量的问题。通过差分思想，有效地解决了区间标记问题。

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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[]) {;}