[vijos P1626] 爱在心中

做完Victoria的舞会3，挑了vijos里强连通分量里面难度值最低的题目，也就是这道。先把第一小问做了，纯Tarjan，只是我学的时候的标程是用邻接表的，这题数据小于是用了邻接矩阵，两者之间的切换花了点时间，我天真地以为i<j等价于i的时间戳小于j的了，呵呵，那时候天真地连d数组都没写…

第二问看别人写了是用n次dfs？我多念了几遍题这**不就是明星奶牛么…感觉这两题基本都没什么差别。放假前老师给明星奶牛的评价是，做了这题图论基本复习了一遍…好吧，是啊，我把书翻出来，敲了一遍拓扑排序…

写完总觉得有点虚，觉得第二问的解法还有漏洞，我对于-1的唯一判断就是拓扑的时候有大于一个的出入度都为零的定点出现，不太清楚这个对不对…这题是不是改一改就能过明星奶牛了？

P.S. 好久没有码100行+的程序了，调试起来非常的虚。是我见识太狭隘还是代码风格太简洁还是做的都是水题？

P.S.2 题目描述里引用了《爱因为在心中》的歌词，挺喜欢这首歌的…~

P.S.3 不要理程序里的那个prim，一脑残忘记拓扑排序叫什么了，不是最小生成树…

program vijos_p1626;
type ty=record
          x,y:integer;
        end;
var map,map2:array[..,..] of word;
    scc,low,s,w,d:array[..] of integer;
    v,sc:array[..] of boolean;
    jl:array[..] of ty;
    n,m,i,j,x,y,top,ans,t,ans2_,ans_t:integer;
procedure tarjan(u:integer);
var i:integer;
begin
  v[u]:=true;
  inc(top);s[top]:=u;sc[u]:=true;
  inc(t);d[u]:=t;low[u]:=t;
  for i:= to n do
    if map[u,i]<> then
      begin
        if v[i]=false then
          begin
            tarjan(i);
            if low[i]<low[u] then low[u]:=low[i];
          end
        else
        if (sc[i]=true) and (d[i]<low[u]) then low[u]:=d[i]; {!!!}
      end;
  if d[u]=low[u] then
    repeat
      i:=s[top];
      scc[i]:=u;
      sc[s[top]]:=false;
      dec(top);
    until i=u;
end;

procedure prim;
var r,c:array[..] of integer;
    stack:array[..] of integer;
    min,min_loc,i,j,top,now:integer;
begin
  fillchar(r,sizeof(r),);
  fillchar(c,sizeof(c),);
  fillchar(stack,sizeof(stack),);
  for i:= to n do
    for j:= to n do
      if (map2[i,j]=) and (i<>j) then
        begin
          inc(r[j]);
          inc(c[i]);
        end;
  top:=;
  for i:= to n do
    if (r[i]=) and (scc[i]=i) then
      begin
        inc(top);
        stack[top]:=i;
        if c[i]= then
          begin
            writeln('-1');
            halt;
          end;
      end;
  while top> do
    begin
      now:=stack[top];dec(top);
      if (c[now]=) and (top<>) then
        begin
          writeln('-1');
          halt;
        end;
      for i:= to n do
        if map2[now,i]= then
          begin
            dec(r[i]);
            if r[i]= then
              begin
                inc(top);
                stack[top]:=i;
              end;
          end;
      if top= then
        begin
          ans2_:=stack[];
          break;
        end;
    end;
end;

begin
  readln(n,m);
  for i:= to m do
    begin
      readln(jl[i].x,jl[i].y);
      map[jl[i].x,jl[i].y]:=;
    end;
  fillchar(v,sizeof(v),false);
  fillchar(w,sizeof(w),);
  for i:= to n do
    if v[i]=false then tarjan(i);
  for i:= to n do
    begin
      inc(w[scc[i]]);
      if w[scc[i]]= then inc(ans);
      if w[scc[i]]= then inc(ans_T);
    end;
  writeln(ans);
  if (ans=) then
    begin
      writeln('-1');
      halt;
    end;
  if (ans=) and (ans_t=) then ans2_:=scc[]
  else
    begin
      for i:= to m do
        map2[scc[jl[i].x],scc[jl[i].y]]:=;   {xiao xin huan}
      ans2_:=-;
      prim;
    end;
  for i:= to n do
    if scc[i]=ans2_ then write(i,' ');
end.

爱在心中

测试数据 #0: Accepted, time = 0 ms, mem = 4704 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 4704 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 4704 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 4704 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 4704 KiB, score = 10

测试数据 #5: Accepted, time = 15 ms, mem = 4704 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 4700 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 4700 KiB, score = 10

测试数据 #8: Accepted, time = 15 ms, mem = 4704 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 4700 KiB, score = 10

Accepted, time = 30 ms, mem = 4704 KiB, score = 100