崇拜有传递性。求所有牛都崇拜的牛
tarjan算法求强连通。
如果不连通就不存在。如果联通,缩点后唯一一个出度为零的点就是答案,有多个则不存在。
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <complex>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cassert>
using namespace std; const int N = 10005;
const int M = 100005; struct Edge {
int to, next;
} edge[M];
int head[N];
int cnt_edge; void add_edge(int u, int v)
{
edge[cnt_edge].to = v;
edge[cnt_edge].next = head[u];
head[u] = cnt_edge;
cnt_edge++;
} int dfn[N];
int low[N];
int stk[N];
bool in[N];
int kind[N]; int top;
int idx;
int cnt; int n, m; void dfs(int u)
{
dfn[u] = low[u] = ++idx;
in[u] = true;
stk[++top] = u;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (!dfn[v])
{
dfs(v);
low[u] = min(low[v], low[u]);
}
else if(in[v])
{
low[u] = min(low[u], dfn[v]);
}
} if (low[u] == dfn[u])
{
++cnt;
int j;
do {
j = stk[top--];
in[j] = false;
kind[j] = cnt;
} while (j != u);
}
} void init()
{
memset(dfn, 0, sizeof dfn);
memset(head, -1, sizeof head);
cnt_edge = 0;
top = cnt = idx = 0;
} int deg[N]; void solve()
{
// 缩点后出度为0的点个数为1
for (int u = 1; u <= n; ++u)
{
int k = kind[u];
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (kind[u] != kind[v]) deg[k]++;
}
} int flag = 0;
int p;
for (int i = 1; i <= cnt; ++i)
{
if (deg[i] == 0)
{
flag++;
p = i;
if (flag > 1) break;
}
}
if (flag == 1)
{
int ans = 0;
for (int i = 1; i <= n; ++i) if (kind[i] == p) ++ans;
printf("%d\n", ans);
}
else
{
printf("0\n");
}
} int main()
{
while (~scanf("%d%d", &n, &m))
{
if (n == 0 && m == 0) break;
int a, b;
init();
for (int i = 0; i < m; ++i)
{
scanf("%d%d", &a, &b);
add_edge(a, b);
} for (int i = 1; i <= n; ++i)
{
if (!dfn[i]) dfs(i);
} solve();
}
return 0;
}