題意:
有n頭牛,m個關係a,b,表示a牛喜歡b牛,現在規定一頭牛是明星的條件是除了他其他所有的牛都喜歡他,現在問你有多少頭牛可以當明星。
思路:
因爲n是1e4,所以floyd求傳遞閉包肯定是不行的,二維數組都開不了這麼大,複雜度也不對,然後就學了下tarjan縮點,因爲每個強連通分量裏的每個牛肯定是互相喜歡的,所以我們要找強連通分量出度爲0的那一團。
1.只找到一團出度爲0的強連通分量,答案就是那一團的大小。
2.找到0或多團強連通分量,答案是0,因爲他們不互相喜歡,也就是不是所有的牛都喜歡他。
#include <bits/stdc++.h>
#define eb emplace_back
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, a, b) for (int i = (int)(a); i >= (int)b; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
#define rep(i, l, r) for (int i = (l); i <= (r); i++)
#define per(i, r, l) for (int i = (r); i >= (l); i--)
#define ms(x, y) memset(x, y, sizeof(x))
#define SZ(x) ((int)(x).size())
using namespace std;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;
typedef pair<i64, i64> pi64;
typedef double ld;
template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }
const int maxn = (int)1e4+1000; //點數
int scc, top, tot;
vector<int> G[maxn];
int low[maxn], dfn[maxn], belong[maxn];
int stk[maxn], vis[maxn];
int sz[maxn], out[maxn];
void init(int n) {
for (int i = 1; i <= n; ++i) {
G[i].clear();
low[i] = 0;
dfn[i] = 0;
stk[i] = 0;
vis[i] = 0;
}
scc = top = tot = 0;
}
void tarjan(int x) {
stk[top++] = x;
vis[x] = 1;
low[x] = dfn[x] = ++tot;
for (int to : G[x]) {
if (!dfn[to]) {
tarjan(to);
low[x] = min(low[x], low[to]);
} else if (vis[to]) {
low[x] = min(low[x], dfn[to]);
}
}
if (low[x] == dfn[x]) {
++scc;
int temp;
do {
temp = stk[--top];
vis[temp] = 0;
belong[temp] = scc;
++sz[scc];
} while (temp != x);
}
}
int n, m;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.precision(10);
cout << fixed;
#ifdef LOCAL_DEFINE
freopen("input.txt", "r", stdin);
#endif
ms(sz, 0);
ms(out, 0);
cin >> n >> m;
init(n);
forn(i, m) {
int u, v;
cin >> u >> v;
G[u].emplace_back(v);
}
for1(i, n) if (!dfn[i]) tarjan(i);
for1(i, n) {
for (int to : G[i]) {
if (belong[i] != belong[to]) ++out[belong[i]];
}
}
int cnt = 0, ans = 0;
for1(i, scc) {
if (!out[i]) {
++cnt;
ans += sz[i];
}
}
if (cnt == 1) cout << ans << '\n';
else cout << 0 << '\n';
#ifdef LOCAL_DEFINE
cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
return 0;
}