POJ 3352

题目大意：给出一副图，求这幅图上可能的生成树中，树的最大边与最小边的差的最小值为多少；

解：枚举最小边，kruskal，然后求出最大的一条边，询问，与答案比较。因为kruskal的最优，所以最小边一定的时候得到的生成树的最大边一定是最小的，注意无解输出1。还有记得并茶几的初始化。

//poj 3352
const
        maxn=1111;
        maxm=1111;
type
        data=record
          next, dest, op: longint;
        end;
var
        edge: array[1..maxn*2]of data;
        deg, pre, dist, low, dfn, vect: array[1..maxn]of longint;
        visit: array[1..maxn]of boolean;
        time, stot, ans, tot, n, m: longint;
procedure add(x, y: longint);
begin
  inc(tot);
  with edge[tot] do begin
    dest := y;
    next := vect[x];
    vect[x] := tot;
    op := tot + 1;
  end;
  inc(tot);
  with edge[tot] do begin
    dest := x;
    next := vect[y];
    vect[y] := tot;
    op := tot - 1;
  end;
end;

procedure init;
var
        i, j, x, y: longint;
begin
  ans := 0; tot := 0;
  fillchar(vect, sizeof(vect), 0);
  readln(n, m);
  for i := 1 to m do begin
    readln(x, y);
    add(x, y);
  end;
end;

procedure tarjan(x: longint);
var
        i: longint;
begin
  inc(time);
  low[x] := time; dfn[x] := time;
  visit[x] := true;
  i := vect[x];
  while i<>0 do
    with edge[i] do begin
     if pre[x]<>dest then
      if dfn[dest]=0 then begin
        pre[dest] := x;
        tarjan(dest);
        pre[dest] := 0;
        if low[dest]<low[x] then low[x] := low[dest];
      end
        else if visit[dest] then begin
          if dfn[dest]<low[x] then low[x] := dfn[dest];
        end;
      i := next;
    end;
  visit[x] := false;
end;

procedure main;
var
        i, u: longint;
begin
  fillchar(dfn, sizeof(dfn), 0);
  fillchar(visit, sizeof(visit), 0);
  time := 0; stot := 0;
  for i := 1 to n do
    if dfn[i]=0 then tarjan(i);
  fillchar(deg, sizeof(deg), 0);
  for u := 1 to n do begin
    i := vect[u];
    while i<>0 do
      with edge[i] do begin
        if low[u]<>low[dest] then begin
          inc(deg[low[u]]);
          inc(deg[low[dest]]);
        end;
        i := next;
      end;
  end;
  for i := 1 to maxn do
    if deg[i] >> 1 = 1 then inc(ans);
  if ans=1 then ans := 0;
  ans := (ans+1)>>1;
end;

procedure print;
begin
  writeln(ans);
end;

begin
  assign(input,'1.txt'); reset(input);
  init;
  main;
  print;
end.