【题目链接】
【算法】
先用tarjan缩点，然后找出度为零的点，即可
【代码】
#include<bits/stdc++.h>
using namespace std;
#define MAXN 500010
struct Edge
{
int to,nxt;
} e[MAXN];
int i,tot,ans,u,v,timer,top,cnt,n,m;
bool instack[MAXN];
int belong[MAXN],low[MAXN],dfn[MAXN],size[MAXN],
head[MAXN],stk[MAXN],x[MAXN],y[MAXN],degree[MAXN];
template <typename T> inline void read(T &x)
{
int f = 1; x = 0;
char c = getchar();
for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
x *= f;
}
template <typename T> inline void write(T x)
{
if (x < 0)
{
putchar('-');
x = -x;
}
if (x > 9) write(x/10);
putchar(x%10+'0');
}
template <typename T> inline void writeln(T x)
{
write(x);
puts("");
}
inline void add(int u,int v)
{
tot++;
e[tot] = (Edge){v,head[u]};
head[u] = tot;
}
inline void tarjan(int u)
{
int i,v;
dfn[u] = low[u] = ++timer;
instack[u] = true;
stk[++top] = u;
for (i = head[u]; i; i = e[i].nxt)
{
v = e[i].to;
if (!dfn[v])
{
tarjan(v);
low[u] = min(low[u],low[v]);
} else
{
if (instack[v])
low[u] = min(low[u],dfn[v]);
}
}
if (dfn[u] == low[u])
{
cnt++;
while (stk[top+1] != u)
{
instack[stk[top]] = false;
belong[stk[top]] = cnt;
size[cnt]++;
top--;
}
}
}
int main() {
read(n); read(m);
for (i = 1; i <= m; i++)
{
read(x[i]); read(y[i]);
add(x[i],y[i]);
}
for (i = 1; i <= n; i++)
if (!dfn[i]) tarjan(i);
for (i = 1; i <= m; i++)
{
if (belong[x[i]] != belong[y[i]])
degree[belong[x[i]]]++;
}
for (i = 1; i <= cnt; i++)
{
if (!degree[i])
{
if (ans)
{
ans = 0;
break;
}
ans = size[i];
}
}
writeln(ans);
return 0;
}