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）

 1 type arr=array[0..100005] of longint;
 2 var
 3    i,j,k,l,m,n,m1,m2:longint;
 4    a:array[0..100005,1..2] of longint;
 5    b,c,d:arr;ans,are:extended;
 6 procedure swap(var x,y:longint);
 7           var z:longint;
 8           begin
 9                z:=x;x:=y;y:=z;
10           end;
11 procedure sort(l,r:longint);
12           var i,j,x,y:longint;
13           begin
14                i:=l;j:=r;x:=a[(l+r) div 2,1];y:=a[(l+r) div 2,2];
15                repeat
16                      while (a[i,1]or ((a[i,1]=x) and (a[i,2]do inc(i);
17                      while (a[j,1]>x) or ((a[j,1]=x) and (a[j,2]>y)) do dec(j);
18                      if i<=j then
19                         begin
20                              swap(a[i,1],a[j,1]);
21                              swap(a[i,2],a[j,2]);
22                              inc(i);dec(j);
23                         end;
24                until i>j;
25                if ithen sort(i,r);
26                if lthen sort(l,j);
27           end;
28 function right(x1,y1,x2,y2:longint):boolean;
29          begin
30               exit((x1*y2)>(x2*y1));
31          end;
32 function trip(x1,y1,x2,y2,x3,y3:longint):boolean;
33          begin
34               exit(right(x2-x1,y2-y1,x3-x2,y3-y2));
35          end;
36 function check(x,y,z:longint):boolean;
37          begin
38               exit(trip(a[x,1],a[x,2],a[y,1],a[y,2],a[z,1],a[z,2]));
39          end;
40 procedure doit(var b:arr;var m:longint);
41           begin
42                b[1]:=d[1];b[2]:=d[2];j:=2;
43                for i:=3 to n do
44                    begin
45                         while (j>1) and not(check(b[j-1],b[j],d[i])) do dec(j);
46                         inc(j);b[j]:=d[i];
47                    end;
48                m:=j;
49           end;
50 begin
51      readln(n);
52      for i:=1 to n do readln(a[i,1],a[i,2]);
53      sort(1,n);j:=1;
54      for i:=2 to n do  //去重
55          begin
56               if (a[i,1]<>a[j,1]) or (a[i,2]<>a[j,2]) then
57                  begin
58                       inc(j);
59                       a[j,1]:=a[i,1];a[j,2]:=a[i,2];
60                  end;
61          end;
62      n:=j;
63     //求凸包
64      for i:=1 to n do d[i]:=i;doit(b,m1);
65      for i:=1 to n do d[i]:=n+1-i;doit(c,m2);
66     //两个半边整合
67      for i:=1 to m1 do d[i]:=b[i];
68      for i:=2 to m2 do d[i+m1-1]:=c[i];
69     //开始计算周长+面积
70      m:=m1+m2-2;ans:=0;are:=0;
71      for i:=1 to m do ans:=ans+sqrt(sqr(a[d[i],1]-a[d[i+1],1])+sqr(a[d[i],2]-a[d[i+1],2]));  //周长
72      for i:=1 to m do are:=are+a[d[i],1]*a[d[i+1],2]-a[d[i],2]*a[d[i+1],1];  //面积
73      are:=abs(are)/2;
74      writeln('Convex Hull:');
75      for i:=1 to m do writeln(a[d[i],1],' ',a[d[i],2]);
76      writeln('C = ',ans:0:1);
77      writeln('S = ',are:0:1);
78      readln;
79 end.

 

转载于:https://www.cnblogs.com/HansBug/p/4442921.html