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 Vi a value Xi (0 ≤ Xi ≤ 1) such that for each edge e(a, b) labeled by op and c, the following formula holds:Xa op Xb = cThe calculating rules are:

Given a Katu Puzzle, your task is to determine whether it is solvable.

  1 #pragma comment(linker, "/STACK:1024000000,1024000000")
  2 #include<iostream>
  3 #include<cstdio>
  4 #include<cstring>
  5 #include<cmath>
  6 #include<math.h>
  7 #include<algorithm>
  8 #include<queue>
  9 #include<set>
 10 #include<bitset>
 11 #include<map>
 12 #include<vector>
 13 #include<stdlib.h>
 14 #include <stack>
 15 using namespace std;
 16 #define PI acos(-1.0)
 17 #define max(a,b) (a) > (b) ? (a) : (b)
 18 #define min(a,b) (a) < (b) ? (a) : (b)
 19 #define ll long long
 20 #define eps 1e-10
 21 #define MOD 1000000007
 22 #define N 1006
 23 #define inf 1e12
 24 int n,m;
 25 vector<int> e[N];
 26
 27 int tot;
 28 int head[N];
 29 int vis[N];
 30 int tt;
 31 int scc;
 32 stack<int>s;
 33 int dfn[N],low[N];
 34 int col[N];
 35 struct Node
 36 {
 37     int from;
 38     int to;
 39     int next;
 40 }edge[N*N];
 41 void init()
 42 {
 43     tot=0;
 44     scc=0;
 45     tt=0;
 46     memset(head,-1,sizeof(head));
 47     memset(dfn,-1,sizeof(dfn));
 48     memset(low,0,sizeof(low));
 49     memset(vis,0,sizeof(vis));
 50     memset(col,0,sizeof(col));
 51 }
 52 void add(int s,int u)//邻接矩阵函数
 53 {
 54     edge[tot].from=s;
 55     edge[tot].to=u;
 56     edge[tot].next=head[s];
 57     head[s]=tot++;
 58 }
 59 void tarjan(int u)//tarjan算法找出图中的所有强连通分支
 60 {
 61     dfn[u] = low[u]= ++tt;
 62     vis[u]=1;
 63     s.push(u);
 64     int cnt=0;
 65     for(int i=head[u];i!=-1;i=edge[i].next)
 66     {
 67         int v=edge[i].to;
 68         if(dfn[v]==-1)
 69         {
 70         //    sum++;
 71             tarjan(v);
 72             low[u]=min(low[u],low[v]);
 73         }
 74         else if(vis[v]==1)
 75           low[u]=min(low[u],dfn[v]);
 76     }
 77     if(dfn[u]==low[u])
 78     {
 79         int x;
 80         scc++;
 81         do{
 82             x=s.top();
 83             s.pop();
 84             col[x]=scc;
 85             vis[x]=0;
 86         }while(x!=u);
 87     }
 88 }
 89 bool two_sat(){
 90
 91     for(int i=0;i<2*n;i++){
 92         if(dfn[i]==-1){
 93             tarjan(i);
 94         }
 95     }
 96     for(int i=0;i<n;i++){
 97         if(col[2*i]==col[2*i+1]){
 98             return false;
 99         }
100     }
101     return true;
102 }
103 int main()
104 {
105     while(scanf("%d%d",&n,&m)==2){
106          init();
107          for(int i=0;i<N;i++) e[i].clear();
108          while(!s.empty()){
109              s.pop();
110          }
111         int a,b,c;
112         char s[6];
113         for(int i=0;i<m;i++){
114             scanf("%d%d%d%s",&a,&b,&c,s);
115             if(s[0]=='A'){
116                 if(c==1){
117                     //e[2*a+1].push_back(2*a);
118                     //e[2*b+1].push_back(2*b);
119                     add(2*a+1,2*a);
120                     add(2*b+1,2*b);
121                 }
122                 else{
123                     //e[2*a].push_back(2*b+1);
124                     //e[2*b].push_back(2*a+1);
125                     add(2*a,2*b+1);
126                     add(2*b,2*a+1);
127                 }
128             }
129             else if(s[0]=='O'){
130                 if(c==1){
131                     //e[2*a+1].push_back(2*b);
132                     //e[2*b+1].push_back(2*a);
133                     add(2*a+1,2*b);
134                     add(2*b+1,2*a);
135                 }
136                 else{
137                     //e[2*a].push_back(2*a+1);
138                     //e[2*b].push_back(2*b+1);
139                     add(2*a,2*a+1);
140                     add(2*b,2*b+1);
141                 }
142             }
143         }
144         if(two_sat()){
145             printf("YES\n");
146         }
147         else{
148             printf("NO\n");
149         }
150     }
151     return 0;
152 }