poj 1797

2013-09-08 09:48

最大生成树，输出生成树中最短的边儿即可

或者对边儿排序，二份答案+BFS判断是否1连通N

时间复杂度都是O(NlogN)的

附最大生成树pascal代码

//By BLADEVIL

var

    m, n, t         :longint;

    pre, other, len :array[0..101000] of longint;

    father          :array[0..100100] of longint;

    i               :longint;

 

procedure init;

var

    i               :longint;

    x, y, z         :longint;

begin

    read(n,m);

    for i:=1 to m do

    begin

        read(x,y,z);

        pre[i]:=x;

        other[i]:=y;

        len[i]:=z;

    end;

    for i:=1 to n do father[i]:=i;

end;

 

procedure qs(low,high:longint);

var

    i, j, x, z      :longint;

begin

    i:=low; j:=high; x:=len[(i+j) div 2];

    while i<j do

    begin

        while x>len[j] do dec(j);

        while x<len[i] do inc(i);

        if i<=j then

        begin

            z:=len[i]; len[i]:=len[j]; len[j]:=z;

            z:=pre[i]; pre[i]:=pre[j]; pre[j]:=z;

            z:=other[i]; other[i]:=other[j]; other[j]:=z;

            inc(i); dec(j);

        end;

    end;

    if i<high then qs(i,high);

    if j>low then qs(low,j);

end;

 

function getfather(x:longint):longint;

begin

    if father[x]=x then exit(x);

    father[x]:=getfather(father[x]);

    exit(father[x]);

end;

 

procedure main;

var

    j               :longint;

    a, b            :longint;

begin

    init;

    qs(1,m);

    for j:=1 to m do

    begin

        a:=getfather(pre[j]);

        b:=getfather(other[j]);

        if a<>b then father[b]:=a;

        a:=getfather(1);

        b:=getfather(n);

        if a=b then

        begin

            writeln('Scenario #',i,':');

            writeln(len[j]);

            writeln;

            exit;

        end;

    end;

end;

 

begin

    read(t);

    for i:=1 to t do main;

end.