POJ 3678 Katu Puzzle （经典2-Sat）

Katu Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6553		Accepted: 2401

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

X ₀ = 1,  X ₁ = 1,  X ₂ = 0,  X ₃ = 1.

Source

POJ Founder Monthly Contest – 2008.07.27, Dagger

 

 

经典2-SAT问题

 

构图时，根据条件找可以确定关系的形如A->B这样的关系式

 

i表示i取1，~i表示i取0

 

i AND j =1    ~i->i, ～j->j, i->j, j->i,后面两个关系式构成一个环，i,j在同一强连通分量中，可以免去

i AND j = 0   i->~i, j->~j 而~i推不出j为0还是1

i OR   j =1    ~i->j, ~j->i

i OR   J =0    i->~i,  j->~j, ~j->~i, ~i->~j 又有环，可以省略

i XOR j =1    i->~j,  j->~i,  ~i->j,  ~j->i

i XOR j =0    i->j,  j->i,  ~i->~j, ~j->~i   又构成两个环

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int VM=2010;
const int EM=4000010;
const int INF=0x3f3f3f3f;

struct Edge{
    int to,nxt;
}edge[EM<<1];

int n,m,cnt,dep,top,atype,head[VM];
int dfn[VM],low[VM],vis[VM],belong[VM];
int stack[VM];

void Init(){
    cnt=0,  atype=0,    dep=0,  top=0;
    memset(head,-1,sizeof(head));
    memset(vis,0,sizeof(vis));
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    memset(belong,0,sizeof(belong));
}

void addedge(int cu,int cv){
    edge[cnt].to=cv;    edge[cnt].nxt=head[cu];     head[cu]=cnt++;
}

void Tarjan(int u){
    dfn[u]=low[u]=++dep;
    stack[top++]=u;
    vis[u]=1;
    for(int i=head[u];i!=-1;i=edge[i].nxt){
        int v=edge[i].to;
        if(!dfn[v]){
            Tarjan(v);
            low[u]=min(low[u],low[v]);
        }else if(vis[v])
            low[u]=min(low[u],dfn[v]);
    }
    int j;
    if(dfn[u]==low[u]){
        atype++;
        do{
            j=stack[--top];
            belong[j]=atype;
            vis[j]=0;
        }while(u!=j);
    }
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d%d",&n,&m)){
        Init();
        char op[10];
        int i,j,c;
        while(m--){
            scanf("%d%d%d%s",&i,&j,&c,op);
            if(op[0]=='A'){
                if(c){
                    addedge(2*i+1,2*i);
                    addedge(2*j+1,2*j);
                    //addedge(2*i,2*j);//2*i和2*j在同一个环中，肯定满足
                    //addedge(2*j,2*i);
                }else{
                    addedge(2*i,2*j+1);
                    addedge(2*j,2*i+1);
                }
            }else if(op[0]=='O'){
                if(c){
                    addedge(2*i+1,2*j);
                    addedge(2*j+1,2*i);
                }else{
                    addedge(2*i,2*i+1);
                    addedge(2*j,2*j+1);
                    //addedge(2*i+1,2*j+1);//同上
                    //addedge(2*j+1,2*i+1);
                }
            }else{
                if(c){
                    addedge(2*i,2*j+1);
                    addedge(2*i+1,2*j);
                    addedge(2*j,2*i+1);
                    addedge(2*j+1,2*i);
                }else{
                    //addedge(2*i,2*j);
                    //addedge(2*j,2*i);
                    //addedge(2*i+1,2*j+1);
                    //addedge(2*j+1,2*i+1);
                }
            }
        }
        for(i=0;i<2*n;i++)
            if(!dfn[i])
                Tarjan(i);
        int flag=1;
        for(i=0;i<n;i++)
            if(belong[2*i]==belong[2*i+1]){
                flag=0;
                break;
            }
        if(flag)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}