UVa 11324 the largest clique/strong connectivity component

There is a direction graph G to find the node set with the largest number of nodes where any 2 points u and v either u can be to V or V can be to u or to each other to reach

First, each of the strongly connected components is accessible, and the weight of each point on a path path is the maximum number of nodes of the strongly connected component, which can be written by a memory search.

#include <cstdio> #include <cstring> #include <vector> #include <stack> using namespace std;
const int MAXN = 1010;
Vector <int> G[MAXN];
Stack <int> s;
int PRE[MAXN];
int LOW[MAXN];
int CNT[MAXN];
int DEGREE[MAXN];
int sccno[maxn];//the scc_cnt int dfs_clock of the strong unicom component of each point;
int scc_cnt;
int n, m;
Vector <int> A[MAXN];

int D[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 (low[u] = = Pre[u]) {scc_cnt++;
			while (1) {int x = S.top ();
			S.pop ();
			SCCNO[X] = scc_cnt;
			cnt[scc_cnt]++;
		if (x = = u) break;
	}}} void Find_scc () {dfs_clock = scc_cnt = 0;
	memset (sccno, 0, sizeof (SCCNO));
	memset (pre, 0, sizeof (pre));
	
for (int i = 1; I <= n; i++) if (!pre[i]) DFS (i); } int DP (int u) {if (D[u]! =-1) return D[u];
	D[u] = Cnt[u];
	int ans = 0;
		for (int i = 0; i < a[u].size (); i++) {int v = a[u][i];
		int temp = DP (v);
	ans = max (ans, temp);
	} D[u] + = ans;
return D[u];
	} int main () {int T;
	scanf ("%d", &t);
		while (t--) {scanf ("%d%d", &n, &m);
			for (int i = 1; I <= n; i++) {g[i].clear ();
		A[i].clear ();
			} for (int i = 1; I <= m; i++) {int u, v;
			scanf ("%d%d", &u, &v);
		G[u].push_back (v);
		} memset (degree, 0, sizeof (degree));
		memset (CNT, 0, sizeof (CNT));
		memset (d,-1, sizeof (d));
		FIND_SCC ();
				for (int u = 1, u <= n; u++) {for (int i = 0; i < g[u].size (); i++) {int v = g[u][i];
					if (sccno[u]! = Sccno[v]) {a[sccno[u]].push_back (sccno[v]);
				degree[sccno[v]]++;
		}}} int ans = 0;
				for (int i = 1; I <= scc_cnt; i++) {if (!degree[i]) {int temp = DP (i);
			ans = max (ans, temp);
	}} printf ("%d\n", ans);
} return 0; }