POJ-2762 Going from u to v or from v to u?

题目大意：

给出一个有向图，这个图，是否存在任意两点a,b可达，这里的任意两点a,b可达是说，只要从a能到b或者只要能从b到a就算是可达的。

解题思路：

先求出这个图的强连通分量，然后缩点建图，只要这个图是一条链状的，那么就可以满足任意两点都可达，否则不满足。

原因是只要这个缩点建图之后的图是链状的，那么必然从链的头到尾，任意两点都可达。一旦不是链状，要么出现分叉，要么某个点入度>=2，这种情况在分叉的两端都是无法可达的。所以只需要是链状就可以了。这个可以用拓扑排序来做，也可以直接一遍dfs求出。

代码：

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

typedef struct node {
	int v, nxt;
	node(int a = 0, int b = 0) {
		v = a; nxt = b;
	}
}Edge;

const int maxn = 1005;
const int maxm = 12010;

int n, m;
stack<int> s;
Edge edge[maxm];
int ComponetNumber;
int tot, head[maxn];
int inComponet[maxn];
vector<int> vec[maxn];
int inx, inStack[maxn];
int low[maxn], dfn[maxn];
vector<int> Componet[maxn];
int in[maxn], out[maxn], vis[maxn];

void init() {
	inx = tot = ComponetNumber = 0;
	while (!s.empty()) s.pop();
	memset(in, 0, sizeof(in));
	memset(dfn, 0, sizeof(dfn));
	memset(low, 0, sizeof(low));
	memset(out, 0, sizeof(out));
	memset(vis, 0, sizeof(vis));
	memset(head, -1, sizeof(head));
	memset(inStack, 0, sizeof(inStack));
	for (int i = 0; i < maxn; ++i) {
		vec[i].clear();
		Componet[i].clear();
	}
}
void add(int u, int v) {
	edge[tot] = Edge(v, head[u]);
	head[u] = tot++;
}
void tarjan(int x) {
	dfn[x] = low[x] = ++inx;
	inStack[x] = 2;
	s.push(x);

	for (int i = head[x]; ~i; i = edge[i].nxt) {
		int v = edge[i].v;
		if (!dfn[v]) {
			tarjan(v);
			low[x] = min(low[x], low[v]);
		} else if (inStack[v] == 2) {
			low[x] = min(low[x], dfn[v]);
		}
	}

	if (dfn[x] == low[x]) {
		++ComponetNumber;
		while(!s.empty()){
			int v = s.top(); s.pop();
			inStack[v] = 1;
			Componet[ComponetNumber].push_back(v);
			inComponet[v] = ComponetNumber;
			if (v == x) break;
		}
	}
}
void dfs(int p) {
	vis[p] = 1;
	int len = vec[p].size();
	for (int i = 0; i < len; ++i) {
		if (!vis[vec[p][i]]) dfs(vec[p][i]);
	}
	if (len > 1) vis[p] = 0;
}
int main() {
	ios::sync_with_stdio(false); cin.tie(0);
	int a, b, t; cin >> t;
	while (t--) {
		cin >> n >> m; init();
		for (int i = 0; i < m; ++i) {
			cin >> a >> b;
			add(a, b);
		}
		for (int i = 1; i <= n; ++i) {
			if (!dfn[i]) tarjan(i);
		}
		for (int i = 1; i <= n; ++i) {
			for (int j = head[i]; ~j; j = edge[j].nxt) {
				int v = edge[j].v;
				if (inComponet[i] != inComponet[v]) {
					++in[inComponet[v]];
					++out[inComponet[i]];
					vec[inComponet[i]].push_back(inComponet[v]);
				}
			}
		}
		for (int i = 1; i <= ComponetNumber; ++i) {
			if (!in[i]) { dfs(i); break; }
		}
		int flag = 1;
		for (int i = 1; i <= ComponetNumber; ++i) {
			if (!vis[i]) flag = 0;
		}
		if (flag) cout << "Yes
";
		else cout << "No
";
	}
	return 0;
}