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

X₀ = 1, X₁ = 1, X₂ = 0, X₃ = 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;
}