poj 3678 Katu Puzzle 2-SAT 有n個變量，每個可以取0或者1，再給出m組關係，每組關係都是兩個變量進行運算可以得到的結果，運算有AND OR XOR三種，問能否滿足這些關係，

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.

 

 //

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=2100000;
int V,E;//點數(1) 邊數
struct edge//鄰接表
{
    int t,w;//u->t=w;
    int next;
};
int p[maxn];//表頭節點
edge G[maxn];
int l;
void init()
{
    memset(p,-1,sizeof(p));
    l=0;
}
//添加邊
void addedge(int u,int t,int w)//u->t=w;
{
    G[l].w=w;
    G[l].t=t;
    G[l].next=p[u];
    p[u]=l++;
}
//tarjan算法 求有向圖強聯通分量
int dfn[maxn],lowc[maxn];
//dfn[u]節點u搜索的次序編號,lowc[u]u或者u的子樹能夠追溯到的棧中的最早的節點
int belg[maxn];//第i個節點屬於belg[i]個強連通分量
int stck[maxn],stop;//stck棧
int instck[maxn];//第i個節點是否在棧中
int scnt;//強聯通分量
int index;
void dfs(int i)
{
    dfn[i]=lowc[i]=++index;
    instck[i]=1;//節點i入棧
    stck[++stop]=i;
    for(int j=p[i];j!=-1;j=G[j].next)
    {
        int t=G[j].t;
        //更新lowc數組
        if(!dfn[t])//t沒有遍歷過
        {
        dfs(t);
        if(lowc[i]>lowc[t]) lowc[i]=lowc[t];
        }//t是i的祖先節點
        else if(instck[t]&&lowc[i]>dfn[t]) lowc[i]=dfn[t];
    }
    //是強連通分量的根節點
    if(dfn[i]==lowc[i])
    {
    scnt++;
    int t;
    do
    {
        t=stck[stop--];
        instck[t]=0;
        belg[t]=scnt;
        }while(t!=i);
    }
}
int tarjan()
{
    stop=scnt=index=0;
    memset(dfn,0,sizeof(dfn));
    memset(instck,0,sizeof(instck));
    for(int i=1;i<=V;i++)
    {
        if(!dfn[i]) dfs(i);
    }
    return scnt;
}
int main()
{
    int n;
    while(scanf("%d%d",&n,&E)==2)
    {
        V=2*n;//1 <=n 0 <=2*n
        init();
        for(int i=0;i<E;i++)
        {
            int u,v,s;
            char str[10];
            scanf("%d%d%d%s",&u,&v,&s,str);u++,v++;
            if(str[0]=='A')
            {
                if(s==1)
                {
                    addedge(u,v,1);
                    addedge(v,u,1);
                    addedge(u+n,u,1);//u,v必選
                    addedge(v+n,v,1);
                }
                else
                {
                    addedge(u,v+n,1);
                    addedge(v,u+n,1);
                }
            }
            else if(str[0]=='O')
            {
                if(s==0)
                {
                    addedge(u+n,v+n,1);
                    addedge(v+n,u+n,1);
                    addedge(u,u+n,1);// !u,!v必選
                    addedge(v,v+n,1);
                }
                else
                {
                    addedge(u+n,v,1);
                    addedge(v+n,u,1);
                }
            }
            else
            {
                if(s==0)
                {
                    addedge(u+n,v+n,1);
                    addedge(v+n,u+n,1);
                    addedge(u,v,1);
                    addedge(v,u,1);
                }
                else
                {
                    addedge(u,v+n,1);
                    addedge(v,u+n,1);
                    addedge(u+n,v,1);
                    addedge(v+n,u,1);
                }
            }
        }
        tarjan();
        int flag=1;
        for(int i=1;i<=n;i++)
        {
            if(belg[i]==belg[i+n])
            {
                flag=0;break;
            }
        }
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}