LA 4287 Proving Equivalences

强连通分量的题，tarjan算法

然后缩点，记a为入度为0的点的个数，b为出度为0的点的个数，ans = max(a, b);

特殊情况缩点后只有一个点 ans = 0;

#include<cstdio>
#include<cstring>
#include<vector>
#include<stack>
#include<algorithm>
using namespace std;

const int MAXN = 20007;
int pt[MAXN], low[MAXN];
int sccn, sccno[MAXN];
int in[MAXN], out[MAXN];
int dt;
vector<int> G[MAXN], scc[MAXN];
stack<int> stk;

int dfs(int u) {
	pt[u] = low[u] = ++dt;
	stk.push(u);
	int i, v, sz = G[u].size();
	for(i = 0; i < sz; i++) {
		v = G[u][i];
		if( !pt[v]) low[u] = min(low[u], dfs(v));
		else if( !sccno[v]) low[u] = min(low[u], pt[v]);
	}
	if(low[u] == pt[u]) {
		sccn ++;
		while( !stk.empty()) {
			v = stk.top(); stk.pop();
			sccno[v] = sccn;
			if(v == u) break;
		}
	}
	return low[u];
}

void getscc(int n) {
	while( !stk.empty()) stk.pop();
	sccn = dt = 0;
	memset(pt, 0 ,sizeof(pt));
	memset(sccno, 0, sizeof(sccno));
	for(int i = 1; i <= n; i++)
		if( !pt[i]) dfs(i);
}

int main() {
	int n, m, T;
	scanf("%d", &T);
	while(T--) {
		scanf("%d%d", &n ,&m);
		int i, j, u, v;
		for(i = 1; i <= n; i++)G[i].clear();
		for(i = 1; i <= m; i++) {
			scanf("%d%d", &u, &v);
			G[u].push_back(v);
		}
		getscc(n);
		memset(in, 0, sizeof(in));
		memset(out, 0, sizeof(out));
		for(i = 1; i <= n; i ++) {
			int sz = G[i].size();
			for(j = 0; j < sz; j++) {
				if( sccno[i] == sccno[G[i][j]]) continue;
				out[sccno[i]] ++;
				in[sccno[G[i][j]]] ++;
			}
		}
		int a = 0, b = 0;
		for(i = 1; i <= sccn; i++) {
			if( !out[i]) a++;
			if( !in[i]) b++;
		}
		int ans = max(a, b);
		if(sccn == 1) ans = 0;
		printf("%d\n", ans);
	}
	return 0;
}