zoukankan      html  css  js  c++  java
  • BZOJ 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚( 线段树 )

    线段树.. 

    --------------------------------------------------------------------------------------

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
     
    #define rep( i , n ) for( int i = 0 ; i < n ; i++ )
    #define clr( x , c ) memset( x , c , sizeof( x ) )
    #define L( x ) ( x << 1 )
    #define R( x ) ( L( x ) ^ 1 )
    #define LC( x ) tree[ L( x ) ]
    #define RC( x ) tree[ R( x ) ]
    #define mid( l , r ) ( ( l + r ) >> 1 )
     
    using namespace std;
     
    const int n = 1000000;
     
    struct Node {
    int l , r;
    int Max , add;
    Node() : Max( 0 ) , add( 0 ) { }
    };
     
    Node tree[ n << 2 ];
     
    void maintain( int x ) {
    Node &o = tree[ x ];
    o.Max = 0;
    if( o.r > o.l )
       o.Max = max( LC( x ).Max , RC( x ).Max );
       
    o.Max += o.add;
    }
     
    int L , R;
     
    void update( int x ) {
    Node &o = tree[ x ];
    if( L <= o.l && o.r <= R ) 
       
       o.add += 1;
       
    else {
    int m = mid( o.l , o.r );
    if( L <= m ) update( L( x ) );
    if( m < R ) update( R( x ) );
    }
    maintain( x );
    }
     
    void build( int x , int l , int r ) {
    Node &o = tree[ x ];
    o.l = l , o.r = r;
    if( l == r )
       return ;
       
    int m = mid( l , r );
    build( L( x ) , l , m );
    build( R( x ) , m + 1 , r );
    int main() {
    // freopen( "test.in" , "r" , stdin );
    int m;
    cin >> m;
    build( 1 , 1 , n );
    while( m-- ) {
    scanf( "%d%d" , &L , &R );
    update( 1 );
    }
    printf( "%d " , tree[ 1 ].Max );
    return 0;
    }

      

    -------------------------------------------------------------------------------------- 

    1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

    Time Limit: 10 Sec  Memory Limit: 64 MB
    Submit: 587  Solved: 327
    [Submit][Status][Discuss]

    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

    Source

  • 相关阅读:
    用java抓取网页信息!
    WPF之Binding对数据的转换(第五天)
    WPF读书笔记 x名称空间详解(第二天)
    WPF的UI布局(Layout)WPF读书笔记(第三天)
    WPF 数据绑定Bingding基础(第四天)
    WPF读书笔记(第一天)
    HelloWin程序(窗口与消息)
    django创建数据库表方法
    OpenSSH利用处理畸形长度密码造成的时间差,枚举系统用户(CVE20166210)
    ldap匿名访问测试脚本
  • 原文地址:https://www.cnblogs.com/JSZX11556/p/4557742.html
Copyright © 2011-2022 走看看