算法模板——计算几何2（二维凸包——Andrew算法）

实现功能：求出二维平面内一对散点的凸包（详见Codevs 1298）

很神奇的算法——先将各个点按坐标排序，然后像我们所知的那样一路左转，求出半边的凸包，然后反过来求另一半的凸包

我以前正是因为总抱着想一步到位的想法，所以每次都跪得很惨（HansBug：事实上这次是我这辈子第一次A掉凸包题）

然后别的没了，就是凸包的基本思想

（顺便输出凸包周长C和面积S）

 type arr=array[..] of longint;
 var
    i,j,k,l,m,n,m1,m2:longint;
    a:array[..,..] of longint;
    b,c,d:arr;ans,are:extended;
 procedure swap(var x,y:longint);
           var z:longint;
           begin
                z:=x;x:=y;y:=z;
           end;
 procedure sort(l,r:longint);
           var i,j,x,y:longint;
           begin
                i:=l;j:=r;x:=a[(l+r) div ,];y:=a[(l+r) div ,];
                repeat
                      while (a[i,]<x) or ((a[i,]=x) and (a[i,]<y)) do inc(i);
                      while (a[j,]>x) or ((a[j,]=x) and (a[j,]>y)) do dec(j);
                      if i<=j then
                         begin
                              swap(a[i,],a[j,]);
                              swap(a[i,],a[j,]);
                              inc(i);dec(j);
                         end;
                until i>j;
                if i<r then sort(i,r);
                if l<j then sort(l,j);
           end;
 function right(x1,y1,x2,y2:longint):boolean;
          begin
               exit((x1*y2)>(x2*y1));
          end;
 function trip(x1,y1,x2,y2,x3,y3:longint):boolean;
          begin
               exit(right(x2-x1,y2-y1,x3-x2,y3-y2));
          end;
 function check(x,y,z:longint):boolean;
          begin
               exit(trip(a[x,],a[x,],a[y,],a[y,],a[z,],a[z,]));
          end;
 procedure doit(var b:arr;var m:longint);
           begin
                b[]:=d[];b[]:=d[];j:=;
                for i:= to n do
                    begin
                         while (j>) and not(check(b[j-],b[j],d[i])) do dec(j);
                         inc(j);b[j]:=d[i];
                    end;
                m:=j;
           end;
 begin
      readln(n);
      for i:= to n do readln(a[i,],a[i,]);
      sort(,n);j:=;
      for i:= to n do  //去重
          begin
               if (a[i,]<>a[j,]) or (a[i,]<>a[j,]) then
                  begin
                       inc(j);
                       a[j,]:=a[i,];a[j,]:=a[i,];
                  end;
          end;
      n:=j;
     //求凸包
      for i:= to n do d[i]:=i;doit(b,m1);
      for i:= to n do d[i]:=n+-i;doit(c,m2);
     //两个半边整合
      for i:= to m1 do d[i]:=b[i];
      for i:= to m2 do d[i+m1-]:=c[i];
     //开始计算周长+面积
      m:=m1+m2-;ans:=;are:=;
      for i:= to m do ans:=ans+sqrt(sqr(a[d[i],]-a[d[i+],])+sqr(a[d[i],]-a[d[i+],]));  //周长
      for i:= to m do are:=are+a[d[i],]*a[d[i+],]-a[d[i],]*a[d[i+],];  //面积
      are:=abs(are)/;
      writeln('Convex Hull:');
      for i:= to m do writeln(a[d[i],],' ',a[d[i],]);
      writeln('C = ',ans::);
      writeln('S = ',are::);
      readln;
 end.