• [HDU] 2795 Billboard [线段树区间求最值]


    Billboard

    Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 11861    Accepted Submission(s): 5223


    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
     
    Sample Output
    1
    2
    1
    3
    -1
     
    Author
    hhanger@zju
     
    Source
     
    Recommend
    lcy
     

    题意:h*w的木板,放进一些1*L的物品,求每次放空间能容纳且最上边的位子
    思路:每次找到最大值的位子,然后减去L
    线段树功能:query:区间求最大值的位子

     

     1 #include<cstdio>
     2 #include<string.h>
     3 #include<algorithm>
     4 
     5 #define clr(x,y) memset(x,y,sizeof(x))
     6 #define lson l,m,rt<<1
     7 #define rson m+1,r,rt<<1|1
     8 
     9 const int N=2e5+3511;
    10 using namespace std;
    11 
    12 int h,w,n,MAX[N<<2];
    13 
    14 void PushUp(int rt)
    15 {
    16     MAX[rt]=max(MAX[rt<<1],MAX[rt<<1|1]);
    17 }
    18 
    19 void build(int l,int r,int rt)
    20 {
    21     int m;
    22     
    23     MAX[rt]=w;
    24     if(l==r) {
    25         return;
    26     }
    27     
    28     m=(l+r)>>1;
    29     build(lson);
    30     build(rson);
    31 }
    32 
    33 int  query(int x,int l,int r,int rt)
    34 {
    35     int m,ret;
    36     if(l==r) {
    37         MAX[rt]-=x;
    38         return l;
    39     }
    40     m=(l+r)>>1;
    41     ret=(MAX[rt<<1]>=x) ? query(x,lson) : query(x,rson);
    42     PushUp(rt);
    43     return ret;
    44 }
    45 
    46 int main()
    47 {
    48     int x;
    49     while(~scanf("%d%d%d",&h,&w,&n)) {
    50         if(h>n) h=n;
    51         build(1,h,1);
    52         while(n--) {
    53             scanf("%d",&x);
    54             if(MAX[1]<x) puts("-1");
    55             else printf("%d
    ",query(x,1,h,1));
    56             
    57         }
    58         
    59     }
    60     
    61     return 0;
    62 }

     

     

  • 相关阅读:
    LinkedBlockingQueue
    PriorityBlockingQueue
    js阻止事件冒泡
    java map常用的4种遍历方法
    JAVA jar 和 war 包的区别
    jquery 操作大全
    Java 使用new Thread和线程池的区别
    原生XMLHttpRequest
    socket 和 webscoket 的区别
    GET和POST请求的区别如下
  • 原文地址:https://www.cnblogs.com/sxiszero/p/4125866.html
Copyright © 2020-2023  润新知