poj3678Katu Puzzle

点为0或1，看满足m个条件时，是否有解

#include<iostream>#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<algorithm>using namespace std;const int MAXN = 1010;  struct twosat  {        int n,c;        vector<int> g[MAXN<<1];        bool mark[MAXN<<1];        int s[MAXN<<1];              bool dfs(int x)       {            int i;          if (mark[x^1])               return 0;            if (mark[x])               return 1;            mark[x] = 1;            s[c++] = x;            for (i = 0; i < g[x].size(); i++)                if (!dfs(g[x][i]))                   return 0;            return 1;        }              void init(int n)       {            int i,t=n<<1;          this->n = n;            for (i = 0; i < t; i++)                 g[i].clear();            memset(mark, 0, sizeof(mark));        }            void add_clause(int x,int xval,int y,int yval)       {        x=x*2+xval;    y=y*2+yval;        g[x].push_back(y^1);            g[y].push_back(x^1);      }            bool solve()       {            int i,t=n<<1;          for (i = 0; i < t; i += 2)                 if (!mark[i] && !mark[i + 1])               {                    c = 0;                    if (!dfs(i))                   {                        while (c > 0)                           mark[s[--c]] =0;                         if (!dfs(i + 1))                           return 0;                    }                }          return 1;        }    }ender;int work(int a,int b,int k){return k==0?a&b:k==1?a|b:a^b;}int main(){char s[100];int a,b,c,n,m,x,y,k;while(cin>>n>>m){ender.init(n);while(m--){scanf("%d%d%d%s",&a,&b,&c,s);k=strcmp(s,"AND")==0?0:strcmp(s,"OR")==0?1:2;for(x=0;x<2;x++)for(y=0;y<2;y++)if(work(x,y,k)!=c)ender.add_clause(a,x,b,y);}cout<<(ender.solve()?"YES":"NO")<<endl;}return 0;}

Katu Puzzle

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8007 Accepted: 2943

Description

Katu Puzzle is presented as a directed graph G(V, E) with each edgee(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integerc (0 ≤ c ≤ 1). One Katu is solvable if one can find each vertexV_i a value X_i (0 ≤ X_i ≤ 1) such that for each edgee(a, b) labeled by op and c, the following formula holds:

 X_a op X_b = c

The calculating rules are:

AND01000101OR01001111XOR01001110

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 40 1 1 AND1 2 1 OR3 2 0 AND3 0 0 XOR

Sample Output

YES

Hint

X₀ = 1, X₁ = 1,X₂ = 0, X₃ = 1.

Source

POJ Founder Monthly Contest – 2008.07.27, Dagger