POJ 3678 Katu Puzzle （2-SAT）

Katu Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5749		Accepted: 2077

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

 

 

比较简单的2-SAT。

给的一些表达式的值，问有没有解。

用2-SAT判定就好了。

/*

POJ 3678

给出两两之间的AND,OR,XOR的值，判断有没有解

典型的2-SAT

*/

#include<stdio.h>

#include<iostream>

#include<algorithm>

#include<vector>

#include<queue>

#include<string.h>

using namespace std;



const int MAXN=2200;//



bool visit[MAXN];

queue<int>q1,q2;

//vector建图方法很妙

vector<vector<int> >adj; //原图    //中间一定要加空格把两个'>'隔开

vector<vector<int> >radj;//逆图

vector<vector<int> >dag;//缩点后的逆向DAG图

int n,m,cnt;



int id[MAXN],order[MAXN],ind[MAXN];//强连通分量，访问顺序，入度



void dfs(int u)

{

    visit[u]=true;

    int i,len=adj[u].size();

    for(i=0;i<len;i++)

      if(!visit[adj[u][i]])

        dfs(adj[u][i]);

    order[cnt++]=u;

}

void rdfs(int u)

{

    visit[u]=true;

    id[u]=cnt;

    int i,len=radj[u].size();

    for(i=0;i<len;i++)

      if(!visit[radj[u][i]])

        rdfs(radj[u][i]);

}

void korasaju()

{

    int i;

    memset(visit,false,sizeof(visit));

    for(cnt=0,i=0;i<2*n;i++)

      if(!visit[i])

        dfs(i);

    memset(id,0,sizeof(id));

    memset(visit,false,sizeof(visit));

    for(cnt=0,i=2*n-1;i>=0;i--)

      if(!visit[order[i]])

      {

          cnt++;//这个一定要放前面来

          rdfs(order[i]);

      }

}

bool solvable()

{

    for(int i=0;i<n;i++)

      if(id[2*i]==id[2*i+1])

        return false;

   return true;

}

int main()

{

    int a,b,c;

    char ch[10];

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        adj.assign(2*n,vector<int>());

        radj.assign(2*n,vector<int>());

        while(m--)

        {

            scanf("%d%d%d%s",&a,&b,&c,&ch);

            int i=a,j=b;

            if(strcmp(ch,"AND")==0)

            {

                if(c==1)//两个都要取1

                {

                    adj[2*i].push_back(2*i+1);

                    adj[2*j].push_back(2*j+1);

                    radj[2*i+1].push_back(2*i);

                    radj[2*j+1].push_back(2*j);

                }

                else //不能两个同时取1

                {

                    adj[2*i+1].push_back(2*j);

                    adj[2*j+1].push_back(2*i);

                    radj[2*j].push_back(2*i+1);

                    radj[2*i].push_back(2*j+1);

                }

            }

            else if(strcmp(ch,"OR")==0)

            {

                if(c==0)//两个都要为0

                {

                    adj[2*i+1].push_back(2*i);

                    adj[2*j+1].push_back(2*j);

                    radj[2*i].push_back(2*i+1);

                    radj[2*j].push_back(2*j+1);

                }

                else

                {

                    adj[2*i].push_back(2*j+1);

                    adj[2*j].push_back(2*i+1);

                    radj[2*j+1].push_back(2*i);

                    radj[2*i+1].push_back(2*j);

                }

            }

            else

            {

                if(c==0)//要相同

                {

                    adj[2*i].push_back(2*j);

                    adj[2*j].push_back(2*i);

                    adj[2*i+1].push_back(2*j+1);

                    adj[2*j+1].push_back(2*i+1);

                    radj[2*i].push_back(2*j);

                    radj[2*j].push_back(2*i);

                    radj[2*i+1].push_back(2*j+1);

                    radj[2*j+1].push_back(2*i+1);

                }

                else

                {

                    adj[2*i].push_back(2*j+1);

                    adj[2*j].push_back(2*i+1);

                    adj[2*i+1].push_back(2*j);

                    adj[2*j+1].push_back(2*i);

                    radj[2*i].push_back(2*j+1);

                    radj[2*j].push_back(2*i+1);

                    radj[2*i+1].push_back(2*j);

                    radj[2*j+1].push_back(2*i);

                }

            }

        }

        korasaju();

        if(solvable())printf("YES\n");

        else printf("NO\n");

    }

    return 0;

}