UVALive - 4287 Proving Equivalences （强连通分量）

链接 ： http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=10294

题意 :  告诉你n个等价的命题 和m个关系 比如 （u，v）代表u可以推导出v， 问至少需要补充多少条边。

用强连通缩点成一张DAG。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <sstream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
#define PII pair<int, int>
#define ALL(x) x.begin(),x.end()
#define mem(a) memset(a,0,sizeof(a))
typedef long long ll;
const double pi = acos(-1.0);
const int MAX = 0x3f3f3f3f;
const ll mod = 1000000007ll;
const int N = 20005;
const int M = 50005;
using namespace std;

int n, m, T;
int he[N];
struct C {
	int ne, to;
} e[M];

void add(int id, int x, int y) {
	e[id].to = y;
	e[id].ne = he[x];
	he[x] = id;
}

stack <int> S;
int pre[N], low[N], scc[N], dfs_clock, scc_cnt;

void dfs(int u) {

	pre[u] = low[u] = ++ dfs_clock;
	S.push(u);
	for(int i = he[u]; i != -1; i = e[i].ne) {
		int v = e[i].to;
		if(pre[v] == 0) {
			dfs(v);
			low[u] = min(low[u], low[v]);
		} else if(scc[v] == 0) {
			low[u] = min(low[u], pre[v]);
		}
	}

	if(low[u] == pre[u]) {
		scc_cnt++;
		for(;;) {
			int x = S.top(); S.pop();
			scc[x] = scc_cnt;
			if(x == u) break;
		}
	}

}

void find_scc() {

	dfs_clock = scc_cnt = 0;
	mem(scc);
	mem(pre);
	for(int i = 1; i <= n; i++) {
		if(pre[i] == 0) {
			dfs(i);
		}
	}

}

int in[N], out[N];

int main() {

	cin >> T;
	while(T--) {

		cin >> n >> m;
		memset(he, -1, sizeof(he));
		for(int i = 1; i <= m; i++) {
			int x, y;
			scanf("%d%d", &x, &y);
			add(i, x, y);
		}

		find_scc();

		for(int i = 1; i <= scc_cnt; i++) {
			in[i] = out[i] = 1;
		}

		for(int u = 1; u <= n; u++) {
			for(int i = he[u]; i != -1; i = e[i].ne) {
				int v = e[i].to;
				if(scc[u] != scc[v]) {
					in[scc[v]] = out[scc[u]] = 0;
				}
			}
		}
		int a = 0, b = 0;
		for(int i = 1; i <= scc_cnt; i++) {
			if(in[i]) a++;
			if(out[i]) b++;
		}

		int ans = max(a, b);
		if(scc_cnt == 1) ans = 0;
		printf("%d\n", ans);

	}

	return 0;
}