[HAOI2006]受欢迎的牛
题目传送门
这道题实质就是缩点，缩完点后一定是$DAG$，统计每个缩点的大小。如果有且只有一个缩点的出度为$0$，说明其他所有点都会直接或间接到达这个点。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define re register 6 #define rep(i, a, b) for (re int i = a; i <= b; ++i) 7 #define repd(i, a, b) for (re int i = a; i >= b; --i) 8 #define maxx(a, b) a = max(a, b); 9 #define minn(a, b) a = min(a, b); 10 #define LL long long 11 #define inf (1 << 30) 12 13 inline int read() { 14 int w = 0, f = 1; char c = getchar(); 15 while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar(); 16 while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar(); 17 return w * f; 18 } 19 20 const int maxn = 1e4 + 5, maxm = 5e4 + 5; 21 22 struct Edge { 23 int u, v, pre; 24 }; 25 26 struct Graph { 27 Edge edges[maxm << 1]; 28 int n, m; 29 int G[maxn]; 30 int dfs_clock, sccno[maxn], scc_cnt, cnt[maxn], pre[maxn], low[maxn]; 31 void init(int n) { 32 this->n = n; 33 m = 0; 34 memset(G, 0, sizeof(G)); 35 dfs_clock = scc_cnt = 0; 36 } 37 void Add(int u, int v) { 38 edges[++m] = (Edge){u, v, G[u]}; 39 G[u] = m; 40 } 41 stack<int> s; 42 void dfs(int u) { 43 pre[u] = low[u] = ++dfs_clock; 44 s.push(u); 45 for (int i = G[u]; i; i = edges[i].pre) { 46 register int v = edges[i].v; 47 if (!pre[v]) { 48 dfs(v); minn(low[u], low[v]); 49 } 50 else if (!sccno[v]) minn(low[u], pre[v]); 51 } 52 if (pre[u] == low[u]) { 53 int x = 0; scc_cnt++; 54 while (x != u) { 55 x = s.top(); s.pop(); 56 sccno[x] = scc_cnt; 57 cnt[scc_cnt]++; 58 } 59 } 60 } 61 } G; 62 63 int n, m, b[maxn]; 64 65 int main() { 66 n = read(), m = read(); 67 G.init(n); 68 69 rep(i, 1, m) { 70 int u = read(), v = read(); 71 G.Add(u, v); 72 } 73 rep(i, 1, n) if (!G.sccno[i]) G.dfs(i); 74 75 rep(i, 1, G.m) { 76 int u = G.edges[i].u, v = G.edges[i].v; 77 if (G.sccno[u] != G.sccno[v]) 78 b[G.sccno[u]] = 1; 79 } 80 81 int ans = 0; 82 rep(i, 1, G.scc_cnt) 83 if (!b[i]) { 84 if (ans) { ans = 0; break; } 85 ans = i; 86 } 87 88 printf("%d", G.cnt[ans]); 89 90 return 0; 91 }
