zoukankan      html  css  js  c++  java
  • POJ 1486

      本来以为水题一笔带过……结果调了大半个小时……

    Sorting Slides
    Time Limit: 1000MS Memory Limit: 10000K
    Total Submissions: 1901 Accepted: 688

    Description

      Professor Clumsey is going to give an important talk this afternoon. Unfortunately, he is not a very tidy person and has put all his transparencies on one big heap. Before giving the talk, he has to sort the slides. Being a kind of minimalist, he wants to do this with the minimum amount of work possible. 
      The situation is like this. The slides all have numbers written on them according to their order in the talk. Since the slides lie on each other and are transparent, one cannot see on which slide each number is written. 

      Well, one cannot see on which slide a number is written, but one may deduce which numbers are written on which slides. If we label the slides which characters A, B, C, ... as in the figure above, it is obvious that D has number 3, B has number 1, C number 2 and A number 4. 

    Your task, should you choose to accept it, is to write a program that automates this process.

    Input

      The input consists of several heap descriptions. Each heap descriptions starts with a line containing a single integer n, the number of slides in the heap. The following n lines contain four integers xmin, xmax, ymin and ymax, each, the bounding coordinates of the slides. The slides will be labeled as A, B, C, ... in the order of the input. 
      This is followed by n lines containing two integers each, the x- and y-coordinates of the n numbers printed on the slides. The first coordinate pair will be for number 1, the next pair for 2, etc. No number will lie on a slide boundary. 
      The input is terminated by a heap description starting with n = 0, which should not be processed. 

    Output

      For each heap description in the input first output its number. Then print a series of all the slides whose numbers can be uniquely determined from the input. Order the pairs by their letter identifier. 
      If no matchings can be determined from the input, just print the word none on a line by itself. 
      Output a blank line after each test case. 

    Sample Input

    4
    6 22 10 20
    4 18 6 16
    8 20 2 18
    10 24 4 8
    9 15
    19 17
    11 7
    21 11
    2
    0 2 0 2
    0 2 0 2
    1 1
    1 1
    0

    Sample Output

    Heap 1
    (A,4) (B,1) (C,2) (D,3)
    
    Heap 2
    none

    【Analysis】

      题目求二分图的必须边,即在二分图G中,i∈G只能与j∈G匹配。

      具体算法也是个条块的东西……

      先求一遍最大匹配,把匹配的结果处理出来,然后枚举每个匹配边,删边,以匹配边得一端增广,若找不到增广路则此为必须边。

      

    program POJ_1486;
    type pos=record
              x1,x2,y1,y2:longint;
         end;
         pos2=record
                   x,y:longint;
         end;
    
    var point:Array[1..1000]of pos;
        num:array[1..1000]of pos2;
        match:array[1..1000]of longint;
        vis:array[1..1000]of boolean;
        n,cases,i,j,u:longint;
        flag:boolean;
        map:array[1..1000,1..1000]of 0..1;
    
    function dfs(u:longint):boolean;
    var i:longint;
    begin
      for i:=1 to n do
        if (not vis[i])and(map[u,i]>0) then
          begin
            vis[i]:=true;
            if (match[i]=-1)or(dfs(match[i])) then
              begin
                match[i]:=u;
                exit(true);
              end;
          end;
      exit(false);
    end;
    
    begin
      Cases:=1;
      readln(n);
      while n<>0 do
        begin
          fillchar(match,sizeof(match),$ff);
          fillchar(map,sizeof(map),0);
          writeln('Heap ',cases);
          inc(cases);
          for i:=1 to n do
            with point[i] do
              readln(x1,x2,y1,y2);
          for i:=1 to n do
            with num[i] do
              readln(x,y);
         for i:=1 to n do
           for j:=1 to n do
             with point[i] do
               if (num[j].x>=x1)and(num[j].x<=x2)and(num[j].y>=y1)and(num[j].y<=y2) then
                 map[j,i]:=1;
         for i:=1 to n do
           begin
             fillchar(vis,sizeof(vis),false);
             dfs(i);
           end;
         flag:=false;
         for i:=1 to n do
           begin
             u:=match[i];
             if u=-1 then continue;
             match[i]:=-1;
             map[u,i]:=0;
             fillchar(vis,sizeof(vis),false);
             if not dfs(u) then
               begin
                 match[i]:=u;
                 if flag then write(' ');
                 flag:=true;
                 write('(',chr(ord('A')+i-1),',',match[i],')');
               end;
             map[u,i]:=1;
           end;
         if not flag then write('none');;
         writeln;writeln;
         readln(n);
        end;
      readln;readln;
    end.
    

      这个题其实不难……但是由于题目没给数据范围,POJ还猥琐地卡10000K内存,我就M了……修正之后又出现了罕见的Presentation error……原来是少输出了一个回车……囧……看来我真的是半秃……

  • 相关阅读:
    JQuery对id中含有特殊字符的转义处理
    jquery 将disabled的元素置为enabled的三种方法
    jeecg表单页面控件权限设置(请先看官方教程,如果能看懂就不用看这里了)
    Google调用explorer.exe打开本地文件
    C++ URLDecode和URLEncode实现——仅限gb2312,非utf8
    jeecg小吐槽续——自己折腾修改在线开发功能中“默认值”的使用
    jeecg小吐槽
    vue使用vant时间日期选择器,日期转化
    vue获取图片宽高
    微信公众号h5用户授权
  • 原文地址:https://www.cnblogs.com/Delostik/p/1989263.html
Copyright © 2011-2022 走看看