Poj 3713 (determines whether an undirected graph is connected in three ways)

Transferring Sylla

Time limit:5000 Ms		Memory limit:65536 K
Total submissions:1160		Accepted:270

Description

After recapturing Sylla, the company plans to establish a new secure system, a transferring net! The new system is designed as follows:

The company staff chooseNCities around the nation which are connected by "Security tunnels" directly or indirectly. once a week, Sylla is to be transferred to another city through the tunnels. as general ordered, the transferring net must reach
A certain security level that there are at least 3 independent paths between any pair of citiesA,B. When General says the paths are independent, he means that the paths share onlyAAndBIn common.

Given a design of a transferring net, your work is to inspect whether it reaches such security level.

Input

The input consists of several test cases.
For each test case, the first line contains two integers,N≤0. 500 andM≤ 20000. indicating the number of cities and tunnels.
The followingMLines each contains two integersAAndB(0 ≤A, B<N), Indicating the cityAAnd CityBAre connected directly by a tunnel.

The input ends by two zeroes.

Output

For each test case output "yes" if it reaches such security level, "no" otherwise.

Sample Input

4 60 10 20 31 21 32 34 50 10 31 21 37 60 10 31 21 32 30 0
 

Sample output

 
Yesnono
 

Source

Poj founder monthly contest-2008.12.28, Dagger question: http://poj.org/problem? Id = 3713 analysis: There is no idea about tri-connectivity. Check that all the solutions of others are enumeration, remove a vertex, and determine whether there is a cut point or is connected. If there is a cut point or not, it is not a three-connection... It's strange. Time is so disgusting. I don't know how these hundreds of MS came from. Code :

# Include <cstdio> using namespace STD; const int Mm = 44444; const int Mn = 555; int T [mm], p [mm]; int H [Mn], dfn [Mn], low [Mn], Du [Mn]; int I, j, k, n, m, idx; bool DFS (INT U, int FA) {dfn [u] = low [u] = ++ idx; For (INT I = H [u], V, son = 0; I> = 0; I = P [I]) if (! Dfn [V = T [I]) {++ son; If (DFS (v, u) return 1; if (FA =-1 & Son> 1 | Fa! =-1 & dfn [u] <= low [v]) return 1; if (low [u]> low [v]) low [u] = low [v];} else if (V! = Fa & low [u]> dfn [v]) low [u] = dfn [v]; return 0;} bool Tarjan () {int I, J, K; for (I = 0; I <n; ++ I) if (Du [I] <3) return 0; for (I = 0; I <n; ++ I) {for (j = idx = 0; j <n; ++ J) dfn [J] = 0; dfn [I] = N + N; for (j = k = 0; j <n; ++ J) if (! Dfn [J]) {If (++ K> 1) return 0; If (DFS (J,-1) return 0 ;}return 1 ;}int main () {While (scanf ("% d", & N, & M), N + M) {for (I = k = 0; I <n; ++ I) H [I] =-1, Du [I] = 0; while (M --) {scanf ("% d", & I, & J ); T [k] = J, P [k] = H [I], H [I] = K ++; t [k] = I, P [k] = H [J], H [J] = K ++; ++ du [I], ++ du [J];} puts (Tarjan ()? "Yes": "no");} return 0 ;}