POJ 3678 Katu Puzzle（2

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:

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

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 ;
}