poj 3678 Katu Puzzle

Katu Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6820		Accepted: 2505

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.

 

题意：一个有向图，给出m个关系a b c op，op是操作种类，有and、or、xor三种，问是否能找出全部结点的值（0或1），使得满足所有的m个关系：a op b=c。

思路：还是2-SAT问题。重要的是如何构图，谨记要连边a->b时，当且仅当选了a就必须选b。所以要这样构图（a表示取0，a+n表示取1）：

op=AND:

c=1: a+n->b+n

         b+n->a+n

         a->a+n      因为这时a和b必须同时为1，a->a+n可以保证a必须为1，下同

         b->b+n

c=0:  a+n->b

          b+n->a

 

op=OR:

c=1: a->b+n

         b->a+n

c=0: a->b

         b->a

         a+n->a

         b+n->b

 

op=XOR:

c=1: a->b+n

         a+n->b

         b->a+n

         b+n->a

c=0: a->b

         a+n->b+n

         b->a

         b+n->a+n

 

 

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <map>
#include <cstdlib>
#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)
#define ll long long
using namespace std;

const int maxn=2005;
const int INF=1000000000;
struct node
{
    int v,next;
}edge[maxn*maxn];
int head[maxn],scc[maxn],stack[maxn];
int low[maxn],dfn[maxn];
bool ins[maxn];
int n,m,num,cnt,top,snum;
void init()
{
    memset(head,-1,sizeof(head));
    num=0;
}
void add(int u,int v)
{
    edge[num].v=v;
    edge[num].next=head[u];
    head[u]=num++;
}
void input()
{
   int a,b,c;
   char op[5];
   while(m--)
   {
       scanf("%d%d%d%s",&a,&b,&c,op);
       if(op[0]=='A')
       {
           if(c==1)
           {
               add(a+n,b+n);
               add(b+n,a+n);
               add(a,a+n);
               add(b,b+n);
           }
           else
           {
               add(a+n,b);
               add(b+n,a);
           }
       }
       else if(op[0]=='O')
       {
           if(c==1)
           {
               add(a,b+n);
               add(b,a+n);
           }
           else
           {
               add(a,b);
               add(b,a);
               add(a+n,a);
               add(b+n,b);
           }
       }
       else
       {
           if(c==1)
           {
               add(a,b+n);
               add(a+n,b);
               add(b,a+n);
               add(b+n,a);
           }
           else
           {
               add(a,b);
               add(a+n,b+n);
               add(b,a);
               add(b+n,a+n);
           }
       }
   }
}
void dfs(int u)
{
    int x;
    dfn[u]=low[u]=++cnt;
    stack[top++]=u;
    ins[u]=true;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(!dfn[v])
        {
            dfs(v);
            low[u]=min(low[u],low[v]);
        }
        else if(ins[v]) low[u]=min(low[u],dfn[v]);
    }
    if(low[u]==dfn[u])
    {
        do{
            x=stack[--top];
            ins[x]=false;
            scc[x]=snum;
        }while(x!=u);
        snum++;
    }
}
void tarjan()
{
    memset(dfn,0,sizeof(dfn));
    memset(ins,false,sizeof(ins));
    cnt=top=snum=0;
    for(int i=0;i<2*n;i++)
    if(!dfn[i]) dfs(i);
}
void solve()
{
    for(int i=0;i<n;i++)
    if(scc[i]==scc[i+n])
    {
        printf("NO\n");
        return;
    }
    printf("YES\n");
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        init();
        input();
        tarjan();
        solve();
    }
    return 0;
}