poj--3678--Katu Puzzle(2-sat 建模)

Katu Puzzle

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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_aopX_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

X ₀ = 1, X ₁ = 1, X ₂ = 0, X ₃ = 1.

#include<stdio.h>
#include<string.h>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
using namespace std;
#define MAX 1000000+10
int low[MAX],dfn[MAX];
int sccno[MAX],m,n;
int scc_cnt,dfs_clock;
bool Instack[MAX];
vector<int>G[MAX];
stack<int>s;
void init()
{
	for(int i=0;i<2*n;i++)
	G[i].clear();
}
void getmap()
{
	while(m--)
	{
		int a,b,c;
		char op[5];
		memset(op,'\0',sizeof(op));
		scanf("%d%d%d%s",&a,&b,&c,op);
		if(op[0]=='A')
		{
			if(c==1)
			{
				G[a+n].push_back(a);
				G[b+n].push_back(b);
			}
			else 
			{
				G[a].push_back(b+n);
				G[b].push_back(a+n);
			}
		}
		else if(op[0]=='O')
		{
			if(c==1)
			{
				G[a+n].push_back(b);
				G[b+n].push_back(a);
			}
			else 
			{
				G[a].push_back(a+n);
				G[b].push_back(b+n);
			}
		}
		else if(op[0]=='X')
		{
			if(c==1)
			{
				G[a+n].push_back(b);
				G[b+n].push_back(a);
				G[a].push_back(b+n);
				G[b].push_back(a+n);
			}
			else 
			{
				G[a].push_back(b);
				G[b].push_back(a);
				G[a+n].push_back(b+n);
				G[b+n].push_back(a+n);
			}
		}
	}
}
void tarjan(int u,int fa)
{
	int v;
	low[u]=dfn[u]=++dfs_clock;
	Instack[u]=true;
	s.push(u);
	for(int i=0;i<G[u].size();i++)
	{
		v=G[u][i];
		if(!dfn[v])
		{
			tarjan(v,u);
			low[u]=min(low[v],low[u]);
		}
		else if(Instack[v])
		low[u]=min(low[u],dfn[v]);
	}
	if(low[u]==dfn[u])
	{
		++scc_cnt;
		for(;;)
		{
			v=s.top();
			s.pop();
			Instack[v]=false;
			sccno[v]=scc_cnt;
			if(v==u) break;
		}
	}
}
void find(int l,int r)
{
	memset(low,0,sizeof(low));
	memset(dfn,0,sizeof(dfn));
	memset(sccno,0,sizeof(sccno));
	memset(Instack,false,sizeof(Instack));
	scc_cnt=dfs_clock=0;
	for(int i=l;i<=r;i++)
	if(!dfn[i]) tarjan(i,-1);
}
void solve()
{
	for(int i=0;i<n;i++)
	{
		if(sccno[i]==sccno[i+n])
		{
			printf("NO\n");
			return ;
		}
	}
	printf("YES\n");
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		init();
		getmap();
		find(0,2*n-1);
		solve();
	}
	return 0;
}