[题解] poj 2762 Going from u to v or from v to u? (tarjan强连通分量)

- 传送门 -

　http://poj.org/problem?id=2762

#Going from u to v or from v to u?

| Time Limit: 2000MS |   | Memory Limit: 65536K |
| Total Submissions: 18189 |   | Accepted: 4868 |

Description

In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn't know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?

Input

The first line contains a single integer T, the number of test cases. And followed T cases.

The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly.

Output

The output should contain T lines. Write 'Yes' if the cave has the property stated above, or 'No' otherwise.

Sample Input

1
3 3
1 2
2 3
3 1

Sample Output

Yes

Source

POJ Monthly--2006.02.26,zgl & twb

- 题意 -

　给一些有向边, 判断图中两点是否联通(从 x 走到 y, 或从 y 走到 x, 满足其中至少一个)
　

- 思路 -

　tarjan缩点后, 我们发现, 一定是一条链才满足条件(两条分岔路上的点无法联通).
　暴力判链...慢如狗...
　注意整张图为一个强连通分量的情况.
　
　细节见代码.
　

- 代码 -

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

const int N = 1e3 + 5;
const int M = 6e3 + 5;

int IN[N], OUT[N];
int STK[N];
int DFN[N], BL[N], LOW[N];
int NXT[M], TO[M], HD[N];
int cas, n, m, sz, cnt, tot, top;
vector<int> G[N];

void init() {
	memset(HD, 0 ,sizeof (HD));
	memset(DFN, 0 ,sizeof (DFN));
	memset(BL, 0, sizeof (BL));
	memset(IN, 0, sizeof (IN));
	memset(OUT, 0, sizeof (OUT));
	sz = cnt = tot = top = 0;
}

void tarjan(int x) {
	DFN[x] = LOW[x] = ++tot;
	STK[++top] = x;
	for (int i = HD[x]; i; i = NXT[i]) {
		int v = TO[i];
		if (!DFN[v]) {
			tarjan(v);
			LOW[x] = min(LOW[x], LOW[v]);
		}
		else if (!BL[v]) LOW[x] = min(LOW[x], DFN[v]);
	}
	if (LOW[x] == DFN[x]) {
		cnt ++;
		int tmp;
		G[cnt].clear();
		do {
			tmp = STK[top--];
			BL[tmp] = cnt;
			G[cnt].push_back(tmp);
		}while (tmp != x);
	}
}

int main() {
	scanf("%d", &cas);
	while (cas--) {
		init();
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= m; ++i) {
			int a, b;
			scanf("%d%d", &a, &b);
			TO[++sz] = b;
			NXT[sz] = HD[a];
			HD[a] = sz;
		}
		for (int i = 1; i <= n; ++i)
			if (!DFN[i]) tarjan(i);
		for (int i = 1; i <= cnt; ++i) {
			for (int j = 0; j < G[i].size(); ++j) {
				int u = G[i][j];
				for (int k = HD[u]; k; k = NXT[k]) {
					int v = TO[k];
					if (i != BL[v]) {
						IN[BL[v]] ++;
						OUT[i] ++;
					}
				}
			}
		}
		int f1 = -1, f2 = -1;
		for (int i = 1; i <= cnt; ++i) {
			if (IN[i] == 0) f1 ++;
			else if (OUT[i] == 0) f2 ++;
			else if (IN[i] != 1 || OUT[i] != 1) {
				f1 = f2 = -2;
				break;
			}
		}
		if ((f1 == 0 && f2 == 0) || cnt == 1) printf("Yes
"); //注意判 cnt==1 的情况
		else printf("No
");
	}
	return 0;
}