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