Poj 3678 Katu puzzle (2-Sat)

Katu puzzle

Time limit:1000 ms		Memory limit:65536 K
Total submissions:5749		Accepted:2077

Description

Katu puzzle is presented as a Directed GraphG(V,E) With each edgeE(A,B) Labeled by a boolean operatorOP(One of and, Or, XOR) and an integerC(0 ≤C≤ 1). One Katu is solvable if one can find each vertexVIA ValueXI(0 ≤XI≤ 1) such that for each edgeE(A,B) LabeledOPAndC, The following formula holds:

 XA OP XB=C

The Calculating Rules are:

And 0 1 0 0 0 1 0 1

Or 0 1 0 0 1 1 1 1

XOR 0 1 0 0 1 1 1 0

Given a Katu puzzle, your task is to determine whether it is solvable.

Input

The first line contains two integersN(1 ≤N≤ 1000) andM, (0 ≤M≤ 1,000,000) indicating the number of vertices and edges.
The followingMLines contain three IntegersA(0 ≤A<N),B(0 ≤B<N),CAnd an operatorOPEach, describing the edges.

Output

Output A line containing "yes" or "no ".

Sample Input

4 40 1 1 and1 2 1 or3 2 0 and3 0 0 XOR
 

Sample output

 
Yes
 

Hint

X 0 = 1, X 1 = 1, X 2 = 0, X 3 = 1.

Source

Poj founder monthly contest-2008.07.27, dagger Relatively simple 2-sat. For the values of some expressions, see if there is any solution. Use 2-sat to determine.

 /*  Poj 3678 provides the and, Or, XOR values between two pairs to determine whether a typical 2-Sat solution is available.  */  # Include <Stdio. h> # Include <Iostream> # Include <Algorithm> # Include <Vector> # Include <Queue> # Include < String . H> Using   Namespace  STD;  Const   Int Maxn = 2200 ; //  Bool Visit [maxn]; queue < Int > Q1, Q2;  //  The method for creating a vector graph is wonderful. Vector <vector < Int > Adj; //  Source image  //  Separate the two '>' with a space in the middle. Vector <vector < Int > Radj; //  Reverse Graph Vector <vector < Int > Dag; //  Reverse Dag after point reduction  Int  N, m, CNT;  Int Id [maxn], order [maxn], IND [maxn]; //  Strongly connected components, access sequence, inbound  Void DFS ( Int  U) {visit [u] = True  ;  Int I, Len = Adj [u]. Size ();  For (I = 0 ; I <Len; I ++ )  If (! Visit [adj [u] [I]) DFS (adj [u] [I]); Order [CNT ++] = U ;}  Void RDFS ( Int  U) {visit [u] = True  ; Id [u] = CNT;  Int I, Len = Radj [u]. Size (); For (I = 0 ; I <Len; I ++ )  If (! Visit [radj [u] [I]) RDFS (radj [u] [I]);}  Void  Korasaju (){  Int  I; memset (visit,  False , Sizeof  (Visit ));  For (CNT = 0 , I =0 ; I < 2 * N; I ++ )  If (! Visit [I]) DFS (I); memset (ID,  0 , Sizeof  (ID); memset (visit,  False , Sizeof  (Visit ));  For (CNT = 0 , I = 2 * N-1 ; I> = 0 ; I -- )  If (! Visit [order [I]) {CNT ++; //  This must be put in front.  RDFS (Order [I]) ;}}  Bool  Solvable (){  For ( Int I = 0 ; I <n; I ++ ) If (ID [ 2 * I] = ID [ 2 * I + 1  ])  Return   False  ;  Return   True  ;}  Int  Main (){  Int  A, B, C;  Char Ch [10  ];  While (Scanf ( "  % D  " , & N, & M )! = EOF) {adj. Assign (  2 * N, vector < Int > (); Radj. Assign (  2 * N, vector < Int > ());  While (M --) {Scanf (  "  % D % s  " , & A, & B, & C ,& Ch );  Int I = A, j = B;  If (Strcmp (CH, "  And  " ) = 0  ){  If (C =1 ) //  Take 1 for both  {Adj [  2 * I]. push_back ( 2 * I + 1  ); Adj [  2 * J]. push_back ( 2 * J + 1  ); Radj [  2 * I + 1 ]. Push_back (2 * I); radj [  2 * J + 1 ]. Push_back ( 2 * J );}  Else   //  You cannot get 1 for both  {Adj [  2 * I + 1 ]. Push_back ( 2 *J); adj [  2 * J + 1 ]. Push_back ( 2 * I); radj [  2 * J]. push_back ( 2 * I + 1  ); Radj [  2 * I]. push_back ( 2 * J + 1  );}} Else   If (Strcmp (CH, "  Or  " ) = 0  ){  If (C = 0 ) //  Both must be 0  {Adj [  2 * I + 1 ]. Push_back ( 2 * I); adj [  2 * J + 1 ]. Push_back ( 2 * J); radj [  2 * I]. push_back ( 2 * I + 1  ); Radj [  2 * J]. push_back ( 2 * J + 1  );} Else  {Adj [  2 * I]. push_back ( 2 * J + 1  ); Adj [  2 * J]. push_back ( 2 * I + 1  ); Radj [  2 * J + 1 ]. Push_back ( 2 *I); radj [  2 * I + 1 ]. Push_back ( 2 * J );}}  Else  {  If (C = 0 ) //  To be the same  {Adj [  2 * I]. push_back ( 2 * J); adj [  2 * J]. push_back ( 2 * I); adj [  2 * I + 1 ]. Push_back ( 2 * J + 1  ); Adj [  2 * J + 1 ]. Push_back ( 2 * I + 1 ); Radj [  2 * I]. push_back ( 2 * J); radj [  2 * J]. push_back ( 2 * I); radj [  2 * I + 1 ]. Push_back ( 2 * J + 1  ); Radj [  2 * J +1 ]. Push_back ( 2 * I + 1  );}  Else  {Adj [  2 * I]. push_back ( 2 * J + 1  ); Adj [  2 * J]. push_back ( 2 * I + 1  ); Adj [ 2 * I + 1 ]. Push_back ( 2 * J); adj [  2 * J + 1 ]. Push_back ( 2 * I); radj [  2 * I]. push_back ( 2 * J + 1  ); Radj [  2 * J]. push_back (2 * I + 1  ); Radj [  2 * I + 1 ]. Push_back ( 2 * J); radj [  2 * J + 1 ]. Push_back ( 2 * I) ;}} korasaju ();  If (Solvable () printf ( " Yes \ n  "  );  Else Printf ( "  No \ n  "  );}  Return   0  ;}