計算幾何&組合數學,也不是很難,關鍵是要有數學儲備
平面向量叉積:a*b=xa*yb-ya*xb;當且僅當a,b共線時取等,a在b右側時爲正,a在b左側時爲負;
剩下的就是用組合數公式一頓算,最後發現只要計算一個點不在另外三個點所成三角形的情況有多少種就能出解,而這個可以枚舉點然後叉積極角排序。
這題在oj上評測很囧,int64超時,longint溢出,real/double才能ac。
貼代碼:
program signaling;
type point=record x,y:double; end;
var x,y:array[1..5000] of double;
a:array[1..5000] of point;
n,i:longint;
sum,ta,tb:double;
function det(a,b:point):double;
begin det:=a.x*b.y-b.x*a.y;
end;
procedure sort(l,r:longint);
var i,j:longint; mid,tmp:point;
begin i:=l; j:=r; mid:=a[(l+r)shr 1];
repeat
while det(a[i],mid)<0do inc(i);
while det(mid,a[j])<0do dec(j);
if i<=j then begin
tmp:=a[i]; a[i]:=a[j]; a[j]:=tmp;
inc(i); dec(j);
end;
until i>j;
if i<r then sort(i,r);
if l<j then sort(l,j);
end;
function c(n,m:int64):double;
var i:double;
begin
if m=2 then c:=n*(n-1)/2;
if m=3 then c:=n*(n-1)*(n-2)/6;
if m=4 then c:=n*(n-1)*(n-2)*(n-3)/24;
end;
procedure process(v:longint);
var i,j:longint;
begin j:=0;
for i:=1 to n do
if i<>v then begin
inc(j);
a[j].x:=x[i]-x[v];
a[j].y:=y[i]-y[v];
end;
sort(1,n-1);
for i:=1 to n-1 do a[i+n-1]:=a[i];
i:=1; j:=2;
while i<=n-1 do begin
while det(a[i],a[j])<0 do inc(j);
if j-1-i>=2 then sum:=sum+c(j-i-1,2);
inc(i);
end;
end;
begin
assign(input,'signaling.in'); reset(input);
assign(output,'signaling.out'); rewrite(output);
readln(n);
for i:=1 to n do readln(x[i],y[i]);
sum:=0;
for i:=1 to n do process(i);
ta:=2*c(n,4)-n*c(n-1,3)+sum;
tb:=c(n,3);
write(ta/tb+3:0:6);
close(input);
close(output);
end.