Ultraviolet A 10510-Cactus (determined by directed cactus chart)

Ultraviolet A 10510-cactus

Question Link

Question: Given a directed graph, ask if this graph is a cactus graph (an edge does not belong to two or more rings)

Idea: similar to constructing DFs of SCC, the determination method is:
1. It must be a strongly connected component
2. A node on a ring must go through only once

In DFS, you only need to record the parent node of each node. If the parent node has been traversed before a node is encountered, you can find and change the node, add the nodes on the ring to + 1 (note that the current node is not counted, because multiple rings can be connected to one node). If one node exceeds 2, it's definitely not a cactus chart.

Code:

#include <cstdio>#include <cstring>#include <vector>#include <stack>#include <algorithm>using namespace std;const int N = 20005;vector<int> g[N];int pre[N], lowlink[N], fa[N], dfs_clock, scc_cnt, cnt[N];int flag;bool find(int v, int u) {while (fa[u] != v) {cnt[u]++;if (cnt[u] > 1)return false;u = fa[u];}return true;}bool dfs_scc(int u) {pre[u] = lowlink[u] = ++dfs_clock;for (int i = 0; i < g[u].size(); i++) {int v = g[u][i];if (!pre[v]) {fa[v] = u;if (!dfs_scc(v)) return false;lowlink[u] = min(lowlink[u], lowlink[v]);} else {lowlink[u] = min(lowlink[u], pre[v]);if (!find(v, u)) return false;}}if (lowlink[u] == pre[u]) {scc_cnt++;if (scc_cnt == 2) return false;}return true;}bool find_scc(int n) {dfs_clock = scc_cnt = 0;memset(cnt, 0, sizeof(cnt));memset(pre, 0, sizeof(pre));for (int i = 0; i < n; i++) {if (!pre[i] && !dfs_scc(i))return false;}return true;}int T, n, m;int main() {scanf("%d", &T);while (T--) {scanf("%d%d", &n, &m);for (int i = 0; i < n; i++)g[i].clear();int u, v;while (m--) {scanf("%d%d", &u, &v);g[u].push_back(v);}printf("%s\n", find_scc(n) ? "YES" : "NO");}return 0;}

Ultraviolet A 10510-Cactus (determined by directed cactus chart)