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

2015-10-11 13:46 447 查看

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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