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.
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.