题目链接：https://www.luogu.org/problemnew/show/P2812
注意：判断出入度是否为0的时候枚举只需到颜色的数量。
坑点：当只有一个强连通分量时，不需要再添加新边。即子任务B ans = 0。
子任务B证明：若每个点都相连通，出入度都必须为1。
保证所有点的出入度都>1就OK。
所以需要找一下出度为0和入度为0的点再取一个max即可。
#include <stack>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100000 + 10;
struct edge{
int from, to, next;
}e[maxn<<2];
int head[maxn], cnt;
bool vis[maxn];
int n, dfn[maxn], low[maxn], tim, color[maxn], num, chudu[maxn], rudu[maxn], runum, chunum;
stack<int> s;
void add(int u, int v)
{
e[++cnt].from = u;
e[cnt].next = head[u];
e[cnt].to = v;
head[u] = cnt;
}
void tarjan(int x)
{
dfn[x] = low[x] = ++tim;
s.push(x); vis[x] = 1;
for(int i = head[x]; i != -1; i = e[i].next)
{
int v = e[i].to;
if(!dfn[v])
{
tarjan(v);
low[x] = min(low[x], low[v]);
}
else if(vis[v])
{
low[x] = min(low[x], low[v]);
}
}
if(dfn[x] == low[x])
{
color[x] = ++num;
vis[x] = 0;
while(s.top() != x)
{
color[s.top()] = num;
vis[s.top()] = 0;
s.pop();
}
s.pop();
}
}
int main()
{
int m;
memset(head, -1, sizeof(head));
scanf("%d",&n);
for(int i = 1; i <= n; i++)
{
int u;
while(scanf("%d",&u) && u != 0)
add(i, u);
}
for(int i = 1; i <= n; i++)
if(!dfn[i]) tarjan(i);
for(int i = 1; i <= n; i++)
for(int j = head[i]; j != -1; j = e[j].next)
{
int v = e[j].to;
if(color[v] != color[i])
{
chudu[color[i]]++;
rudu[color[v]]++;
}
}
for(int i = 1; i <= num; i++)
{
if(!rudu[i]) runum++;
if(!chudu[i]) chunum++;
}
for(int i = 2; i <= n; i++)
{
if(color[i] != color[i-1])
{
printf("%d\n%d",runum, max(runum, chunum));
return 0;
}
}
printf("%d\n%d",runum,0);
return 0;
}