强连通分量缩点 出度为0并且唯一一个强连通分量的大小是所求的答案 不唯一输出0
#include#include #include #include #include using namespace std; const int maxn = 10010; vector G[maxn]; int pre[maxn]; int low[maxn]; int sccno[maxn]; int dfs_clock; int scc_cnt; stack S; int n, m; int degree[maxn]; int cnt[maxn]; void dfs(int u) { pre[u] = low[u] = ++dfs_clock; S.push(u); for(int i = 0; i < G[u].size(); i++) { int v = G[u][i]; if(!pre[v]) { dfs(v); low[u] = min(low[u], low[v]); } else if(!sccno[v]) low[u] = min(low[u], pre[v]); } if(pre[u] == low[u]) { scc_cnt++; while(1) { cnt[scc_cnt]++; int x = S.top(); S.pop(); sccno[x] = scc_cnt; if(x == u) break; } } } void find_scc() { dfs_clock = scc_cnt = 0; memset(sccno, 0, sizeof(sccno)); memset(pre, 0, sizeof(pre)); memset(cnt, 0, sizeof(cnt)); for(int i = 1; i <= n; i++) if(!pre[i]) dfs(i); } int main() { while(scanf("%d %d", &n, &m) != EOF) { for(int i = 1; i <= n; i++) G[i].clear(); while(m--) { int u, v; scanf("%d %d", &u, &v); G[u].push_back(v); } find_scc(); memset(degree, 0, sizeof(degree)); for(int i = 1; i <= n; i++) { for(int j = 0; j < G[i].size(); j++) { int v = G[i][j]; if(sccno[i] != sccno[v]) { degree[sccno[i]]++; } } } int flag = 0; int p = 1; for(int i = 1; i <= scc_cnt; i++) { if(degree[i] == 0) { flag++; p = i; } } if(flag > 1) printf("0\n"); else printf("%d\n", cnt[p]); } return 0; }