回复 @noip : 附议，Cogs ac在洛谷只有55分，

回复 @noip : 附议，Cogs ac在洛谷只有55分，

tarjan

COGS的淼数据没有“有多个SCC的出度都为0”的数据，所以不判断符合要求的SCC是否唯一也可以AC，然而POJ是有这样的数据的...

VIP蒟蒻的百题纪念。

数据弱QwQ这能过的代码POJ 2186死活不过

邻接矩阵开不了的统计奇技淫巧
证明:缩点后的图中有出度的点不是受欢迎的。
假设它受欢迎，则所有点到它都有边，所以它和它的出点互达。因此不是DAG，与题设矛盾。
立得受欢迎的点出度为0
然而这说明出度为0是必要条件而不是充分条件。
所以要判断有几个这样的点。

在 poj 上 过不了 wa
啊啊 求助 poj 2186

回复 @◆半城烟沙灬��你打天下 :
~~~装逼遭雷劈！！！~~~

回复 @FF_Sky||幻 : VIP这TM的是填空题么？？？

VIP上面的2b不要再刷了，2不2啊

#include<cstring>

#include<iostream>

#include<cstdio>

using namespace std;

#define __________________ 10005

#define ___________________ 50005

struct node{

 int __________, ___________;

}________________[___________________];

int _________________[__________________],____[___________________];

int _[__________________],______[__________________],_____[__________________],___[__________________],__[__________________];

bool ____________________[__________________];

int ______________;

char _______________;

int n,m,s,_______,___s,________,_________;

void _____________(int __________,int ___________){

 ________________[++s].__________=__________;

 ________________[s].___________=___________;

 ____[s]=_________________[__________];

 _________________[__________]=s;

}

void ____________(int x){

 int i;

 _______++;

 _[x]=______[x]=_______;

 _____[++________]=x;

 ____________________[x]=1;

 for (i=_________________[x];i;i=____[i]){

 if (!_[________________[i].___________]){

 ____________(________________[i].___________);

 ______[x]=min(______[x],______[________________[i].___________]);

 }

 else{

 if (____________________[x]){

 ______[x]=min(______[x],_[________________[i].___________]);

 }

 }

 }

 if (______[x]==_[x]){

 ___s++;

 while (_____[________+1]!=x){

 ___[_____[________]]=___s;

 __[___s]++;

 ____________________[_____[________]]=0;

 ________--;

 }

 }

}

void _____________________(){

 int i;

 memset(____________________,1,sizeof(____________________));

 for (i=1;i<=m;i++){

 if (___[________________[i].__________]==___[________________[i].___________]) continue;

 ____________________[___[________________[i].__________]]=0;

 }

}

int ______________________(){

 while ((_______________=getchar()) && (_______________>'9' || _______________<'0'));

 ______________=_______________-'0';

 while ((_______________=getchar()) && (_______________<='9' && _______________>='0')){

 ______________=______________*10+_______________-'0';

 }

 return ______________;

}

int main(){

 freopen("cow.in","r",stdin);

 freopen("cow.out","w",stdout);

 int i,__________,___________;

 n=______________________();

 m=______________________();

 s=0;

 for (i=1;i<=m;i++){

 __________=______________________();

 ___________=______________________();

 _____________(__________,___________);

 }

 _______=0;

 ________=0;

 ___s=0;

 for (i=1;i<=n;i++){

 if (_[i]) continue;

 ____________(i);

 }

 _____________________();

 _________=0;

 for (i=1;i<=___s;i++){

 if (____________________[i]){

 if (_________) {

 cout<<0<<endl;

 return 0;

 }

 _________=__[i];

 }

 }

 cout<<_________<<endl;

 return 0;

}

回复 @LuciFer_T-J :
~~~装逼遭雷劈！！！~~~

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
struct sky
{
int ff,tt,next;
};
sky c[50005];
int _[10005],__[10005],___[10005],____[10005],_____[10005];
int ______[50005];
bool v[50005];
int n,m,top,now,colors,ans,tot,ret;
char ch;
inline void add(int x,int y)
{
tot++;
c[tot].ff=x;
c[tot].tt=y;
c[tot].next=______[x];
______[x]=tot;
}
inline int read()
{
while (!isdigit(ch = getchar()));
ret = ch-48;
while (isdigit(ch = getchar())) (ret *= 10) += ch-48;
return ret;
}
void tarjan(int x)
{
now++;
_[x]=__[x]=now;
___[++top]=x;
v[x]=1;
for (int i=______[x];i;i=c[i].next)
{
if (!_[c[i].tt])
{
tarjan(c[i].tt);
__[x]=min(__[x],__[c[i].tt]);
}
else
{
if (v[x])
{
__[x]=min(__[x],_[c[i].tt]);
}
}
}
if (__[x]==_[x])
{
colors++;
while (___[top+1]!=x)
{
____[___[top]]=colors;
_____[colors]++;
v[___[top]]=0;
top--;
}
}
}
int main()
{
freopen("cow.in","r",stdin);
freopen("cow.out","w",stdout);
n = read(); m = read();
memset(______,0,sizeof(______));
tot=0;
for (int i=1;i<=m;i++)
{
int x,y;
x=read();y=read();
add(x,y);
}
now=top=colors=0;
memset(_____,0,sizeof(_____));
memset(_,0,sizeof(_));
memset(v,0,sizeof(v));
for (int i=1;i<=n;i++)
{
if (_[i]) continue;
tarjan(i);
}
memset(v,1,sizeof(v));
for (int i=1;i<=tot;i++)
{
if (____[c[i].ff]==____[c[i].tt]) continue;
v[____[c[i].ff]]=0;
}
ans=0;
for (int i=1;i<=colors;i++)
{
if (v[i])
{
if (ans)
{
printf("0");
return 0;
}
ans=_____[i];
}
}
printf("%d",ans);
}

这成了机房里的竞速题了- -

数据太弱了。。。我的明显错误做法都能过

因为入度会有重边，所以要求出度。

一开始居然把分支内部的边也算到它的入度里了……智商是硬伤……