[2-SAT] poj 3678 Katu Puzzle

题目链接：

Katu Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7888		Accepted: 2888

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

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题目意思：

给一幅有向图，每条边代表连接的两个节点的一种运算，并且告诉运算后的值，问能否有一种节点取值，使得各边都满足边的运算。

解题思路：

2-SAT

i代表1 ~i带表0

当i&j==1时，建边 ~i->i ~j->j

当i&j==0时，建边 i->~j j->~i ~i->~j ~j->~i

当i|j==1时，建边 ~i->j ~j->i

当i|j==0时，建边 i->~i j->~j

当i^j==1时，建边 i->~j ~i->j j->~i ~j->i

当i^j==0时，建边 i->j ~i->~j j->i ~j->~i

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 2200

int n,m;
vector<vector<int> >myv;
int low[Maxn],dfn[Maxn],sc,bc,dep;
int in[Maxn],sta[Maxn];
bool iss[Maxn];

void tarjan(int cur)
{
    int ne;
    low[cur]=dfn[cur]=++dep;
    sta[++sc]=cur;
    iss[cur]=true;

    for(int i=0;i<myv[cur].size();i++)
    {
        ne=myv[cur][i];
        if(!dfn[ne])
        {
            tarjan(ne);
            low[cur]=min(low[cur],low[ne]);
        }
        else if(iss[ne]&&dfn[ne]<low[cur])
            low[cur]=dfn[ne];
    }
    if(low[cur]==dfn[cur])
    {
        ++bc;
        do
        {
            ne=sta[sc--];
            in[ne]=bc;
            iss[ne]=false;
        }while(ne!=cur);
    }
}
void solve()
{
    sc=bc=dep=0;

    memset(dfn,0,sizeof(dfn));
    memset(iss,false,sizeof(iss));
    for(int i=0;i<2*n;i++)
        if(!dfn[i])
            tarjan(i);
}

int main()
{
    //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   while(~scanf("%d%d",&n,&m))
   {
       myv.clear();
       myv.resize(2*n+10);

       for(int i=1;i<=m;i++)
       {
           int a,b,c;
           char ss[10];
           scanf("%d%d%d%s",&a,&b,&c,ss);
           if(*ss=='A')
           {
               if(c==1)
               {
                   myv[2*a].push_back(2*a+1);
                   myv[2*a+1].push_back(2*b+1);
                   myv[2*b].push_back(2*b+1);
                   myv[2*b+1].push_back(2*a+1);
               }
               else
               {
                   myv[2*a+1].push_back(2*b);
                   myv[2*b+1].push_back(2*a);
               }
           }
           else if(*ss=='O')
           {
               if(c==1)
               {
                   myv[2*a].push_back(2*b+1);
                   myv[2*b].push_back(2*a+1);
               }
               else
               {
                   myv[2*a+1].push_back(2*a);
                   myv[2*a].push_back(2*b);
                   myv[2*b].push_back(2*a);
                   myv[2*b+1].push_back(2*b);
               }
           }
           else
           {
               if(c==1)
               {
                   myv[2*a].push_back(2*b+1);
                   myv[2*a+1].push_back(2*b);
                   myv[2*b].push_back(2*a+1);
                   myv[2*b+1].push_back(2*a);
               }
               else
               {
                   myv[2*a].push_back(2*b);
                   myv[2*a+1].push_back(2*b+1);
                   myv[2*b].push_back(2*a);
                   myv[2*b+1].push_back(2*a+1);
               }
           }
       }
       solve();
       bool ans=true;

       for(int i=0;i<n;i++)
       {
           if(in[2*i]==in[2*i+1])
           {
               ans=false;
               break;
           }
       }
       if(ans)
            printf("YES\n");
       else
            printf("NO\n");
   }
    return 0;
}

时间： 2024-12-21 17:46:35

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