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  • SSL 1720 Surround the Trees

    Description

      There are a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the minimal required length of the rope. However, he does not know how to calculate it. Can you help him?
      The diameter and length of the trees are omitted, which means a tree can be seen as a point. The thickness of the rope is also omitted which means a rope can be seen as a line.
    这里写图片描述
      There are no more than 100 trees.

    Input

      The input contains one or more data sets. At first line of each input data set is number of trees in this data set, it is followed by series of coordinates of the trees. Each coordinate is a positive integer pair, and each integer is less than 32767. Each pair is separated by blank.

      Zero at line for number of trees terminates the input for your program.

    Output

     The minimal length of the rope. The precision should be 10^-2.
    

    题解

     找到最右下角的点,然后暴力枚举其他(n-1)个点。
    

    代码

    type
      arr=record
            x,y:longint;
          end;
    var
      n,top:longint;
      ans:real;
      e:array [0..501] of arr;
      s:array [0..501] of longint;
    function cj(p1,p2,p0:arr):longint;
    begin
      cj:=(p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
    end;
    
    function comp(p1,p2:arr):boolean;
    var
      t:longint;
    begin
      t:=cj(p1,p2,e[0]);
      if (t>0) or (t=0) and (sqr(p1.x-e[0].x)+sqr(p1.y-e[0].y)<sqr(p2.x-e[0].x)+sqr(p2.y-e[0].y)) then
        exit(true);
      exit(false);
    end;
    
    procedure qsort(l,r:longint);
    var
      i,j:longint;
      mid,t:arr;
    begin
      if l>=r then exit;
      mid:=e[l+random(r-l+1)];
      i:=l; j:=r;
      repeat
        while comp(e[i],mid) do inc(i);
        while comp(mid,e[j]) do dec(j);
        if i<j then
          begin
            t:=e[i]; e[i]:=e[j]; e[j]:=t;
          end;
      until i>=j;
      qsort(l,j);
      qsort(j+1,r);
    end;
    
    procedure init;
    var
      i:longint;
      t:arr;
    begin
      randomize;
      for i:=0 to n-1 do
        begin
          readln(e[i].x,e[i].y);
          if (e[i].y<e[0].y) or (e[i].y=e[0].y) and (e[i].x<e[0].x) then
            begin
              t:=e[i]; e[i]:=e[0]; e[0]:=t;
            end;
        end;
      qsort(1,n-1);
    end;
    
    procedure main;
    var
      i:longint;
    begin
      for i:=1 to 3 do
        s[i]:=i-1;
      top:=3;
      for i:=3 to n-1 do
        begin
          while cj(e[i],e[s[top]],e[s[top-1]])>=0 do dec(top);
          inc(top);
          s[top]:=i;
         end;
      ans:=0;
      if n>1 then
        begin
          s[top+1]:=s[1];
          for i:=1 to top do
            ans:=ans+sqrt(sqr(e[s[i]].x-e[s[i+1]].x)+sqr(e[s[i]].y-e[s[i+1]].y));
        end;
      if n=2 then ans:=sqrt(sqr(e[1].x-e[0].x)+sqr(e[1].y-e[0].y))*2;
      writeln(ans:0:2);
    end;
    
    begin
      readln(n);
      while n<>0 do
        begin
          fillchar(e,sizeof(e),0);
          fillchar(s,sizeof(s),0);
          init;
          main;
          readln(n);
        end;
    end.
    
    
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  • 原文地址:https://www.cnblogs.com/zyx-crying/p/9319573.html
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