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  • 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[]) {;}
     
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  • 原文地址:https://www.cnblogs.com/ZlycerQan/p/7409790.html
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