POJ 3678 Katu Puzzle（2 - SAT） - from lanshui_Yang

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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_i a 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:

AND01000101OR01001111XOR01001110

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 40 1 1 AND1 2 1 OR3 2 0 AND3 0 0 XOR

Sample Output

YES

Hint

X₀ = 1, X₁ = 1, X₂ = 0, X₃ = 1.

题目大意：有 n 个变量和m个条件，每个条件格式如下：

给你三个数a ， b ， c 和一个运算字符串op (AND , OR , XOR 等)，要求Xa op Xb = c 。

问：是否能给每个变量赋值（0 或 1），使得每个条件都能满足？

解题思路：这道题是一道2 - SAT问题的变形，2 - SAT 问题中是通过条件建立图中的边，这道题也类似，也是也要通过条件建立相应的边，只不过有些细节需要注意。

请看代码：

#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<vector>#include<cmath>#include<cstdio>#include<queue>#define mem(a , b) memset(a , b , sizeof(a))using namespace std ;const int MAXN = 10000 ;vector<int> G[MAXN * 2] ;bool mark[MAXN * 2] ;int S[MAXN] , c ;  // 模拟栈char op[8] ;int n , m ;int pan ;  // 判断标志void chu(){    int i ;    for(i = 0 ; i < n * 2 ; i ++)        G[i].clear() ;    mem(mark , 0) ;}void init(){    chu() ;    pan = 0 ;    int i ;    for(i = 0 ; i < m ; i ++)    {        int a , b , c ;        scanf("%d%d%d" , &a , &b , &c) ;        scanf("%s" , op) ;        if(op[0] == 'A')        {            if(c == 1)  // 注意此时建边的方式            {                G[2 * a].push_back(2 * a + 1) ;                  G[2 * b].push_back(2 * b + 1) ;            }            else            {                G[2 * a + 1].push_back(2 * b) ;                G[2 * b + 1].push_back(2 * a) ;            }        }        else if(op[0] == 'X')        {            if(c == 0)            {                G[2 * a].push_back(2 * b) ;                G[2 * a + 1].push_back(2 * b + 1) ;                G[2 * b].push_back(2 * a) ;                G[2 * b + 1].push_back(2 * a + 1) ;            }            else            {                G[2 * a].push_back(2 * b + 1) ;                G[2 * a + 1].push_back(2 * b) ;                G[2 * b].push_back(2 * a + 1) ;                G[2 * b + 1].push_back(2 * a) ;            }        }        else        {            if(c == 0) // 注意此时建边的方式            {                G[2 * a + 1].push_back(2 * a) ;                G[2 * b + 1].push_back(2 * b) ;            }            else            {                G[2 * a].push_back(2 * b + 1) ;                G[2 * b].push_back(2 * a + 1) ;            }        }    }}bool dfs(int x){    if(mark[x ^ 1]) return false ;    if(mark[x]) return true ;    mark[x] = true ;    S[c ++] = x ;    int i ;    for(i = 0 ; i < G[x].size() ; i ++)    {        if(!dfs(G[x][i]))            return false ;    }    return true ;}void solve(){    if(pan)        puts("NO") ;    else    {        int i ;        for(i = 0 ; i < n ; i ++)        {            if(!mark[i * 2] && !mark[i * 2 + 1])            {                c = 0 ;                if(!dfs(i * 2))                {                    while (c > 0)                    {                        mark[ S[-- c] ] = false ;                    }                    if(!dfs(i * 2 + 1))                    {                        pan = 1 ;                        break ;                    }                }            }        }        if(pan)            puts("NO") ;        else            puts("YES") ;    }}int main(){    while (scanf("%d%d" , &n , &m) != EOF)    {        init() ;        solve() ;    }    return 0 ;}