2-sat（and,or,xor）poj3678

Katu Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7949		Accepted: 2914

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_ia 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

题意：给出一个连通图对于每条边都有一种操作（and，or，xor），使两个端点的操作结果是c，问是否存在这样一个连通图，存在输出YES否者输出NO
分析：裸的2-sat操作
公式：i&j=1  (i-->i+n)  (j-->j+n)
           i&j=0  (i+n-->j)  (j+n-->i)
           i|j=1   (i-->j+n)  (j-->i+n)
           i|j=0   (i+n-->i)  (j+n-->j)
          i^j=1   (i-->j+n)  (j+n-->i)  (j-->i+n)  (i+n-->j)
          i^j=0   (i-->j)  (j-->i)  (i+n-->j+n)  (j+n-->i+n)
程序;
#include"string.h"
#include"stdio.h"
#include"iostream"
#include"algorithm"
#include"queue"
#include"stack"
#include"stdlib.h"
#include"math.h"
#define inf 10000000
#define INF 0x3f3f3f3f
const double PI=acos(-1.0);
const double r2=sqrt(2.0);
const int M=1010;
const int N=1010*1000*4;
#define eps 1e-10
using namespace std;
struct node
{
    int u,v,next;
}edge[N];
stack<int>q;
int t,head[M],dfn[M],low[M],belong[M],use[M],num,index;
void init()
{
    t=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
    edge[t].u=u;
    edge[t].v=v;
    edge[t].next=head[u];
    head[u]=t++;
}
void tarjan(int u)
{
    dfn[u]=low[u]=++index;
    q.push(u);
    use[u]=1;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(!dfn[v])
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(use[v])
            low[u]=min(low[u],dfn[v]);
    }
    if(low[u]==dfn[u])
    {
        num++;
        int p;
        do
        {
            p=q.top();
            q.pop();
            use[p]=0;
            belong[p]=num;
        }while(p!=u);
    }
}
int slove(int n)
{
    num=index=0;
    memset(use,0,sizeof(use));
    memset(dfn,0,sizeof(dfn));
    for(int i=0;i<n*2;i++)
        if(!dfn[i])
        tarjan(i);
    for(int i=0;i<n;i++)
        if(belong[i]==belong[i+n])
        return 0;
    return 1;
}
int main()
{
    int n,m,a,b,c,i;
    char str[9];
    while(scanf("%d%d",&n,&m)!=-1)
    {
        init();
        for(i=1;i<=m;i++)
        {
            scanf("%d%d%d%s",&a,&b,&c,str);
            if(strcmp(str,"AND")==0)
            {
                if(c)
                {
                    add(a,a+n);
                    add(b,b+n);
                }
                else
                {
                    add(b+n,a);
                    add(a+n,b);
                }
            }
            else if(strcmp(str,"OR")==0)
            {
                if(c)
                {
                    add(b,a+n);
                    add(a,b+n);
                }
                else
                {
                    add(a+n,a);
                    add(b+n,b);
                }
            }
            else
            {
                if(c)
                {
                    add(a,b+n);
                    add(b,a+n);
                    add(b+n,a);
                    add(a+n,b);
                }
                else
                {
                    add(a,b);
                    add(b,a);
                    add(a+n,b+n);
                    add(b+n,a+n);
                }
            }
        }
        if(slove(n))
            printf("YES\n");
        else
            printf("NO\n");
    }
}