• Ikki's Story IV


    题意/Description:

        liympanda, one of Ikki’s friend, likes playing games with Ikki. Today after minesweeping with Ikki and winning so many times, he is tired of such easy games and wants to play another game with Ikki.

        liympanda has a magic circle and he puts it on a plane, there are n points on its boundary in circular border: 0, 1, 2, …, n − 1. Evil panda claims that he is connecting m pairs of points. To connect two points, liympanda either places the link entirely inside the circle or entirely outside the circle. Now liympanda tells Ikki no two links touch inside/outside the circle, except on the boundary. He wants Ikki to figure out whether this is possible…

        Despaired at the minesweeping game just played, Ikki is totally at a loss, so he decides to write a program to help him.

     

    读入/Input

        The input contains exactly one test case.

        In the test case there will be a line consisting of of two integers: n and m (n ≤ 1,000, m ≤ 500). The following m lines each contain two integers ai and bi, which denote the endpoints of the ith wire. Every point will have at most one link.

     

    输出/Output

        Output a line, either “panda is telling the truth...” or “the evil panda is lying again”.

    题解/solution


    看了图,就知道了要把每条边看成2个点:分别表示在内部连接和在外部连接,只能选择一个。计作点i和点i'。然后思考一下怎么连边:

    1、i->j'表示如果i在里,j'在外

    2、j->i' , i'->j , j'->i 同理

    之后用2-SAT做,用tarjan找i和i'是不是强连通,如果不是,则成立。

     

    代码/Code

    <span style="font-family:SimSun;"><strong>const
      maxE=10000001;
      maxV=100001;
    type
      arr=record
        x,y,next:longint;
      end;
    var
      n,m,nm,op,num:longint;
      tu:array [0..maxE] of arr;
      v:array [0..maxE] of boolean;
      ls,tx,ty,sta,bel,dfn,low:array [0..maxV] of longint;
    procedure add(o,p:longint);
    begin
      inc(nm);
      with tu[nm] do
        begin
          x:=o; y:=p;
          next:=ls[o];
          ls[o]:=nm;
        end;
    end;
    
    procedure init;
    var
      i,j,k:longint;
    begin
      readln(n,m);
      for i:=1 to m do
        begin
          readln(tx[i],ty[i]);
          if tx[i]>ty[i] then
            begin
              k:=tx[i]; tx[i]:=ty[i]; ty[i]:=k;
            end;
        end;
      nm:=0;
      for i:=1 to m do
        for j:=i+1 to m do
          if (tx[j]>tx[i]) and(tx[j]<ty[i]) and (ty[j]>ty[i]) or (ty[j]>tx[i]) and (ty[j]<ty[i]) and (tx[j]<tx[i]) then
            begin
              add(i+m,j);
              add(j,i+m);
              add(i,j+m);
              add(j+m,i);
            end;
    end;
    
    function min(t,k:longint):longint;
    begin
      if t<k then exit(t);
      exit(k);
    end;
    
    procedure tarjan(o:longint);
    var
      i,k,tmp:longint;
    begin
      inc(num);
      dfn[o]:=num; low[o]:=num;
      inc(op);
      sta[op]:=o; v[o]:=true;
      i:=ls[o];
      while i<>0 do
        begin
          if dfn[tu[i].y]=0 then
            begin
              tarjan(tu[i].y);
              low[o]:=min(low[o],low[tu[i].y]);
            end else
              if v[tu[i].y] then low[o]:=min(low[o],dfn[tu[i].y]);
          i:=tu[i].next;
        end;
      if dfn[o]=low[o] then
        begin
          inc(k);
          tmp:=-1;
          while tmp<>o do
            begin
              tmp:=sta[op];
              dec(op);
              bel[tmp]:=k;
              v[tmp]:=false;
            end;
        end;
    end;
    
    procedure print;
    var
      i:longint;
      boo:boolean;
    begin
      for i:=1 to m+m do
        if dfn[i]=0 then tarjan(i);
      boo:=true;
      for i:=1 to m do
        if bel[i]=bel[i+m] then boo:=false;
      if boo then write('panda is telling the truth...')
             else write('the evil panda is lying again');
    end;
    
    begin
      init;
      print;
    end.
    </strong></span>


  • 相关阅读:
    IOS 面试题(不断更新...)
    IOS 数字日期转化为字符串
    C#汉字生成简拼
    ObjectiveC 深浅拷贝
    数组遍历方法forEach 和 map 的区别
    COJ1174(Shining Gems)
    POJ1062(昂贵的聘礼)
    HDOJ1879(继续畅通工程)
    最短路径经典题集(转载)
    HDOJ1863(畅通工程)
  • 原文地址:https://www.cnblogs.com/zyx-crying/p/9319683.html
Copyright © 2020-2023  润新知