POJ 3678 Katu Puzzle（2 - SAT） - from lanshui_Yang
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 Vi a value Xi (0 ≤ Xi ≤ 1) such that for each edge e(a, b) labeled by op and c, the following formula holds:
Xa op Xb = c
The calculating rules are:
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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
#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 ;
}