Katu Puzzle (poj 3678 2-SAT)

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Katu Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8487		Accepted: 3149

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

题意：n个点给出m个关系（AND,OR,XOR），问是否存在解。

思路：２-SAT建图

a AND b = 1: ~x->x,~y->y （两个数必须全为1）
a AND b = 0: y->~x,x->~y （两个数至少有一个为0）
a OR b = 1：~x->y,~y->x （两个数至少有一个为1）
a OR b = 0: x->~x,y->~y （两个数必须全为0）
a XOR b = 1:x->~y,y->~x,~y->x,~x->y （两个数必须不同）
a XOR b = 0:x->y,y->x,~x->~y,~y->~x （两个数必须相同）

代码：

#include <iostream>
#include <functional>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 3005;
const int MAXM = 1200010;
const int N = 1005;

struct Edge
{
    int to,next;
}edge[MAXM];

int tot,head[MAXN];
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
bool Instack[MAXN];
int top,Index,scc;
int n,m;
char str[10];

void init()
{
    tot=0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v)
{
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void Tarjan(int u)
{
    int v;
    Low[u]=DFN[u]=++Index;
    Stack[top++]=u;
    Instack[u]=true;
    for (int i=head[u];~i;i=edge[i].next)
    {
        v=edge[i].to;
        if (!DFN[v])
        {
            Tarjan(v);
            if (Low[u]>Low[v])
                Low[u]=Low[v];
        }
        else if (Instack[v]&&Low[u]>DFN[v])
            Low[u]=DFN[v];
    }
    if (Low[u]==DFN[u])
    {
        ++scc;
        do{
            v=Stack[--top];
            Instack[v]=false;
            Belong[v]=scc;
        }while (v!=u);
    }
    return ;
}

bool solvable(int n)
{
    memset(DFN,0,sizeof(DFN));
    memset(Instack,false,sizeof(Instack));
    Index=scc=top=0;
    for (int i=0;i<n;i++)
        if (!DFN[i])
            Tarjan(i);
    for (int i=0;i<n;i+=2)
    {
        if (Belong[i]==Belong[i^1])
            return false;
    }
    return true;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    int i,j,a,b,c;
    scanf("%d%d",&n,&m);
    init();
    for (i=0;i<m;i++)
    {
        scanf("%d%d%d %s",&a,&b,&c,str);
        if (strcmp(str,"OR")==0)
        {
            if (c==0)
            {
                addedge(2*a,2*a+1);
                addedge(2*b,2*b+1);
            }
            else
            {
                addedge(2*a+1,2*b);
                addedge(2*b+1,2*a);
            }
        }
        else if (strcmp(str,"AND")==0)
        {
            if (c==0)
            {
                addedge(2*a,2*b+1);
                addedge(2*b,2*a+1);
            }
            else
            {
                addedge(2*a+1,2*a);
                addedge(2*b+1,2*b);
            }
        }
        else
        {
            if (c==0)
            {
                addedge(2*a,2*b);
                addedge(2*a+1,2*b+1);
                addedge(2*b,2*a);
                addedge(2*b+1,2*a+1);
            }
            else
            {
                addedge(2*a,2*b+1);
                addedge(2*a+1,2*b);
                addedge(2*b,2*a+1);
                addedge(2*b+1,2*a);
            }
        }
    }
    if (solvable(2*n))
        printf("YES\n");
    else
        printf("NO\n");
    return 0;
}