[BZOJ1626] [Usaco2007 Dec]Building Roads 修建道路

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传送门

http://www.lydsy.com/JudgeOnline/problem.php?id=1626

题目大意

给你n个点的坐标,m条边所连的两点,输出最小连通代价

题解

最小生成树,把已经连上的点先连上,注意算两点间距离爆longint

var fa:array[0..1000]of longint; x:array[0..1000,0..1000]of longint; y:array[0..1000,1..2]of longint; z:array[0..2000000]of double; t:array[0..2000000,1..2]of longint; i,j,k:longint; n,m:longint; a,b,len:longint; ans:double;procedure sort(l,r:longint);var i,j,c:longint; a,b:double;begin i:=l; j:=r; a:=z[(l+r) div 2]; repeat  while z[i]<a do inc(i);  while a<z[j] do dec(j);  if not(i>j) then   begin    b:=z[i]; z[i]:=z[j]; z[j]:=b;    c:=t[i,1]; t[i,1]:=t[j,1]; t[j,1]:=c;    c:=t[i,2]; t[i,2]:=t[j,2]; t[j,2]:=c;    inc(i); dec(j);   end; until i>j; if l<j then sort(l,j); if i<r then sort(i,r);end;function get(a:longint):longint;begin if fa[a]=a then exit(a); fa[a]:=get(fa[a]); exit(fa[a]);end;begin readln(n,m); for i:=1 to n do  readln(y[i,1],y[i,2]); fillchar(x,sizeof(x),0); for i:=1 to n do  fa[i]:=i; for i:=1 to m do  begin readln(a,b); x[a,b]:=1; fa[get(a)]:=get(b); end; len:=0; for i:=1 to n do  for j:=1 to n do   begin    if (i=j)or(x[i,j]=1) then continue;    inc(len); z[len]:=sqrt(int64(y[i,1]-y[j,1])*int64(y[i,1]-y[j,1])+int64(y[i,2]-y[j,2])*int64(y[i,2]-y[j,2]));    t[len,1]:=i; t[len,2]:=j;   end; sort(1,len); {z[i]} ans:=0; for i:=1 to len do  begin   a:=get(t[i,1]); b:=get(t[i,2]);   if a<>b then begin ans:=ans+z[i]; fa[a]:=b; end;  end; writeln(ans:0:2);end.

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