POJ 3678 Katu Puzzle 2-SAT-优快云博客

本文链接：https://blog.youkuaiyun.com/aosakixuan/article/details/45787927

本文介绍了一种解决KatuPuzzle问题的算法实现，该问题被定义为在一个带有布尔运算符标记的有向图中寻找每个节点的可行值。文章提供了完整的代码示例，解释了如何通过构建图并应用Tarjan算法来确定谜题是否可解。

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Katu Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8330		Accepted: 3069

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

WA了几发之后我想明白了一个问题。。然后过了>.<

/** Author: ¡î¡¤aosaki(*¡¯(OO)¡¯*)  niconiconi¡ï **/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
//#include <tuple>
#define ALL(v) (v).begin(),(v).end()
#define foreach(i,v) for (__typeof((v).begin())i=(v).begin();i!=(v).end();i++)
#define SIZE(v) ((int)(v).size())
#define mem(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define lp(k,a) for(int k=1;k<=a;k++)
#define lp0(k,a) for(int k=0;k<a;k++)
#define lpn(k,n,a) for(int k=n;k<=a;k++)
#define lpd(k,n,a) for(int k=n;k>=a;k--)
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d %d",&a,&b)
#define lowbit(x) (x&(-x))
#define ll long long
#define pi pair<int,int>
#define vi vector<int>
#define PI acos(-1.0)
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define TT cout<<"*****"<<endl;
#define TTT cout<<"********"<<endl;
inline int gcd(int a,int b)
{
    return a==0?b:gcd(b%a,a);
}

#define INF 1e9
#define eps 1e-8
#define mod 10007
#define maxn 5010
#define maxm 4000010
using namespace std;

int v[maxn],col[maxn];
int rea[maxn],low[maxn],stack[maxn];
int n,m,tot=0,color,tm=0,top=0;
int pre[maxn];

struct Side
{
    int to,next;
}e[maxm];

void add(int u,int v)
{
    e[tot].to=v;
    e[tot].next=pre[u];
    pre[u]=tot++;
}

void tarjan(int i)
{
    v[i]=1;
    top++;
    stack[top]=i;
    ++tm;
    rea[i]=tm;
    low[i]=tm;
    for(int j=pre[i];j!=-1;j=e[j].next)
        {
            int x=e[j].to;
            if(v[x]==0) tarjan(x);
            if(v[x]<2) low[i]=min(low[i],low[x]);
        }
    if(rea[i]==low[i])
    {
        color++;
        while(stack[top+1]!=i)
        {
            col[stack[top]]=color;
            v[stack[top]]=2;
            top--;
        }

    }

}


void init()
{
    mem(col);
    mem(v);
    mem1(pre);
    tot=0;
    top=0;
    tm=0;
    color=0;
    mem(low);
    mem(rea);
    mem(stack);
}


bool ok()
{
    lp(i,n)
    {
        if((col[i*2]==col[i*2+1]))
            return 0;
    }
    return 1;
}

int main()
{
     freopen("in.txt","r",stdin);
     int a,b,c;
     char s[10];
     while(~sc2(n,m))
     {
        init();
        lp(i,m)
        {
            sc(a);sc(b);sc(c);
            scanf("%s",s);
            a++; b++;
            if(s[0]=='A')
            {
                 if(c==1)
                {
                    add(a*2,a*2+1);
                    add(b*2,b*2+1);
                }
                else
                {
                    add(a*2+1,b*2);
                    add(b*2+1,a*2);
                }
            }

            if(s[0]=='O')
            {
                if(c==1)
                {
                    add(a*2,b*2+1);
                    add(b*2,a*2+1);
                }
                else
                {
                    add(a*2+1,a*2);
                    add(b*2+1,b*2);
                  
                }

            }

            if(s[0]=='X')
            {
                if(c==1)
                {
                    add(a*2+1,b*2);
                    add(b*2+1,a*2);
                    add(a*2,b*2+1);
                    add(b*2,a*2+1);
                }
                else
                {
                    add(a*2+1,b*2+1);
                    add(b*2+1,a*2+1);
                    add(b*2,a*2);
                    add(a*2,b*2);
                }
            }

        }
        lp(i,n*2)
          if(!rea[i])
             tarjan(i);
        if(ok()) printf("YES\n");
          else printf("NO\n");

     }

    return 0;
}