Time to live

Http://acm.hust.edu.cn: 8080/judge/problem/viewproblem. Action? Id = 24028

In fact, the principles of BFS and DFS are the same. They all find the deepest, but they both need to be done twice.

Find the deepest depth for the first time, and find the deepest depth for the second time

View code

# Include <iostream> # Include < String . H> # Include <Stdio. h> # Include <Algorithm> # Include <Vector> # Include <Queue> # Define INF ~ 0u> 1 # Define Maxn 100000 Using   Namespace  STD;  Int  N; vector < Int > Node [maxn]; pair < Int , Int > Ans;  Int  Visit [maxn];  Void  Init (){  Int A, B;  For ( Int I = 0 ; I <n; I ++ ) Node [I]. Clear ();  For ( Int I = 1 ; I <n; I ++ ) {CIN > A> B; node [A]. push_back (B); node [B]. push_back ();}}  Void BFS ( Int  Root) {queue <Pair < Int , Int > Q; //  Queue with two attributes Q. Push (make_pair (root, 0 )); //  First, enter the root team Ans. Second = 0  ; Memset (visit,  0 , Sizeof  (Visit); visit [root] = 1  ; While (! Q. Empty () {pair < Int , Int > Temp = Q. Front (); q. Pop ();  If (Temp. Second> Ans. Second) {ans. Second = Temp. Second; ans. First = Temp. First ;}  For ( Int I = 0 ; I <node [temp. First]. Size (); I ++ )  If (! Visit [node [temp. First] [I]) //  No access to the team  {Visit [node [temp. First] [I] = 1  ; Q. Push (make_pair (node [temp. First] [I], temp. Second + 1  ));}}}  Int  Main (){  Int  Test; For (CIN> test; test -- ) {CIN > N; Init (); BFS (  0 ); //  For the first time, assume that 0 is the root of the BFS, and find the deepest point in depth.  BFS (ANS. First );  //  The second time, looking for the deepest point in Depth          If (ANS. Second % 2  ) Cout <Ans. Second/ 2 +1 <Endl; //  Half of the even answers plus 1          Else  Cout <Ans. Second/ 2 <Endl; //  Half of odd answers  }  Return   0  ;}