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.