2-sat（and,or,xor）poj3678

Katu Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7949		Accepted: 2914

Description

Katu Puzzle is presented as a directed graph G(V, E) with each edge e(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integer c (0 ≤ c ≤ 1). One Katu is solvable if one can find each vertex V_ia value X_i (0 ≤ X_i≤ 1) such that for each edge e(a, b) labeled by op and c, the following formula holds:

X_a op X_b = 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 integers N (1 ≤ N ≤ 1000) and M,(0 ≤ M ≤ 1,000,000) indicating the number of vertices and edges.
The following M lines contain three integers a (0 ≤ a < N), b(0 ≤ b < N), c and an operator op each, describing the edges.

Output

Output a line containing "YES" or "NO".

Sample Input

4 4
0 1 1 AND
1 2 1 OR
3 2 0 AND
3 0 0 XOR

Sample Output

YES


题意：给出一个连通图对于每条边都有一种操作（and，or，xor），使两个端点的操作结果是c，问是否存在这样一个连通图，存在输出YES否者输出NO
分析：裸的2-sat操作
公式：i&j=1  (i-->i+n)  (j-->j+n)
           i&j=0  (i+n-->j)  (j+n-->i)
           i|j=1   (i-->j+n)  (j-->i+n)
           i|j=0   (i+n-->i)  (j+n-->j)
          i^j=1   (i-->j+n)  (j+n-->i)  (j-->i+n)  (i+n-->j)
          i^j=0   (i-->j)  (j-->i)  (i+n-->j+n)  (j+n-->i+n)
程序;