• HDU-------(2795)Billboard(线段树区间更新)


    Billboard

    Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 10594    Accepted Submission(s): 4686


    Problem Description
    At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is the place where all possible announcements are posted: nearest programming competitions, changes in the dining room menu, and other important information.

    On September 1, the billboard was empty. One by one, the announcements started being put on the billboard.

    Each announcement is a stripe of paper of unit height. More specifically, the i-th announcement is a rectangle of size 1 * wi.

    When someone puts a new announcement on the billboard, she would always choose the topmost possible position for the announcement. Among all possible topmost positions she would always choose the leftmost one.

    If there is no valid location for a new announcement, it is not put on the billboard (that's why some programming contests have no participants from this university).

    Given the sizes of the billboard and the announcements, your task is to find the numbers of rows in which the announcements are placed.
     
    Input
    There are multiple cases (no more than 40 cases).

    The first line of the input file contains three integer numbers, h, w, and n (1 <= h,w <= 10^9; 1 <= n <= 200,000) - the dimensions of the billboard and the number of announcements.

    Each of the next n lines contains an integer number wi (1 <= wi <= 10^9) - the width of i-th announcement.
     
    Output
    For each announcement (in the order they are given in the input file) output one number - the number of the row in which this announcement is placed. Rows are numbered from 1 to h, starting with the top row. If an announcement can't be put on the billboard, output "-1" for this announcement.
     
    Sample Input
    3 5 5 2 4 3 3 3
     
    Sample Output
    1 2 1 3 -1
     
    Author
    hhanger@zju
     
    Source

     代码:     代码写的比较脆,花掉的时间是:3795ms 呵~~呵~~ ,呵了个呵的!

     1 /*基础的线段树*/
     2 #include<cstdio>
     3 #include<cstring>
     4 
     5 const int maxn= 200050;
     6 int aa[maxn];
     7 
     8 struct node
     9 {
    10     int lef,rig,mx;
    11     int mid(){
    12       return lef+((rig-lef)>>1);
    13     }
    14 };
    15 
    16 int h,w,n;
    17 node  reg[maxn<<2];
    18 inline int max(int a,int b)
    19 {
    20   return a>b?a:b;
    21 }
    22 
    23 void Build(int lef,int rig,int pos)
    24 {
    25      reg[pos]=(node){lef,rig,w};
    26      if(lef==rig) return ;
    27      int mid=reg[pos].mid();
    28      Build(lef,mid,pos<<1);
    29      Build(mid+1,rig,pos<<1|1);
    30 }
    31 
    32 void Work(int val,int pos)
    33 {
    34     if(reg[pos].mx>=val)
    35     {
    36        if(reg[pos].rig==reg[pos].lef)
    37        {
    38           reg[pos].mx-=val;
    39           printf("%d
    ",reg[pos].rig);
    40           return ;
    41        }
    42        if(val<=reg[pos<<1].mx)
    43            Work(val,pos<<1);
    44        else 
    45            Work(val,pos<<1|1);
    46       reg[pos].mx=max(reg[pos<<1].mx,reg[pos<<1|1].mx);
    47      }
    48    else if(pos==1) printf("-1
    ");
    49    return ;
    50 }
    51 
    52 int main()
    53 {
    54     int cn;
    55    while(scanf("%d%d%d",&h,&w,&n)!=EOF)
    56    {
    57         if(h>n)  h=n;
    58          Build(1,h,1);
    59       while(n--)
    60       {
    61         scanf("%d",&cn);
    62         Work(cn,1);
    63       }
    64    }
    65    return 0;
    66 }
    View Code
  • 相关阅读:
    redis-3.2.5 make 报错
    haproxy 实现多域名证书https
    centos7修改主机名
    ngx_image_thumb模块生成缩略图
    查看nginx在安装时开启了哪些模块
    C# install-package:"xx"已拥有为“xxx”定义的依赖项
    JetBrains 2017/2018全系列产品激活工具
    查看win10版本方法,及win10升级方法
    Windows 10正式版的历史版本
    open '/dev/hwlog_switch' fail -1, 13. Permission denied
  • 原文地址:https://www.cnblogs.com/gongxijun/p/3895885.html
Copyright © 2020-2023  润新知