bzoj 1626[Usaco2007 Dec]Building Roads 修建道路

来源：互联网发布：罗德岛大学怎么样知乎编辑：程序博客网时间：2024/05/21 03:17

题意：n个点，m条已有的边，建一些新边，代价为两点间的欧几里得距离，使得n个点联通，求最少的代价

...高仿kruscal..

先把已经连上的连上，然后再把没有连上的边从小到大排序，按kruscal的套路去跑就行了（详见代码）

注意：再求欧几里得距离的时候，由于x和y的范围为10^6，会爆int，要强转long long（int64）

type        rec=record            a,b:longint;            len:double;end;var        n,m,tot,tt,tx,ty:longint;        xx,yy           :longint;        i,j             :longint;        l               :array[0..1000010] of rec;        x,y,father      :array[0..1010] of longint;        ans             :double;function get_father(x:longint):longint;begin   if x=father[x] then exit(x);   father[x]:=get_father(father[x]);   exit(father[x]);end;procedure sort(ll,rr:longint);var        i,j:longint;        x:double;        y:rec;begin   i:=ll; j:=rr; x:=l[(ll+rr)>>1].len;   while (i<=j) do   begin      while l[i].len<x do inc(i);      while l[j].len>x do dec(j);      if (i<=j) then      begin         y:=l[i]; l[i]:=l[j]; l[j]:=y;         inc(i); dec(j);      end;   end;   if i<rr then sort(i,rr);   if j>ll then sort(ll,j);end;begin   read(n,m);   for i:=1 to n do father[i]:=i;   for i:=1 to n do read(x[i],y[i]);   for i:=1 to m do   begin      read(xx,yy);      tx:=get_father(xx);      ty:=get_father(yy);      if tx<>ty then      begin         father[tx]:=ty; inc(tt);      end;   end;   //   for i:=1 to n do     for j:=i+1 to n do       if get_father(i)<>get_father(j) then       begin          inc(tot);          l[tot].a:=i; l[tot].b:=j;          l[tot].len:=sqrt(int64(x[i]-x[j])*int64(x[i]-x[j])+int64(y[i]-y[j])*int64(y[i]-y[j]));       end;   sort(1,tot);   ans:=0;   for i:=1 to tot do     if tt=n-1 then break else     begin         tx:=get_father(l[i].a);         ty:=get_father(l[i].b);         if tx<>ty then         begin            father[tx]:=ty;            inc(tt);            ans:=ans+l[i].len;         end;      end;   writeln(ans:0:2);end.

——by Eirlys

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