题意：有n个顶点里面可以放数字1或0，给m个限制，每个限制给出两个顶点编号和两编号内数字运算后的结果

思路：很直接的2-SAT，每个点分为1和0两种情况，按限制要求建边，跑tarjan然后判断点是否在同一个强连通分量里就OK了

(一下代码是WA的。。找了一晚找不到BUG)

#include
const int maxn = 1e3+;
int stack[maxn],dfn[maxn<<],low[maxn<<],head[maxn*maxn],dfs_num,top;
int color[maxn<<],col_num;
bool vis[maxn*maxn];
class edge
{
public:
int to,next;
}e[];
inline int gmin(int a,int b)
{
return a<b?a:b;
}
int ans;
void addedge(int u,int v)
{
e[++ans].next=head[u];
e[ans].to=v;
head[u]=ans;
}
void Tarjan ( int x ) {
dfn[ x ] = ++dfs_num ;
low[ x ] = dfs_num ;
vis [ x ] = true ;//是否在栈中
stack [ ++top ] = x ;
for ( int i=head[ x ] ; i!= ; i=e[i].next ){
int temp = e[ i ].to ;
if ( !dfn[ temp ] ){
Tarjan ( temp ) ;
low[ x ] = gmin ( low[ x ] , low[ temp ] ) ;
}
else if ( vis[ temp ])low[ x ] = gmin ( low[ x ] , dfn[ temp ] ) ;
}
if ( dfn[ x ]==low[ x ] ) {//构成强连通分量
vis[ x ] = false ;
color[ x ] = ++col_num ;//染色
while ( stack[ top ] != x ) {//清空
color [stack[ top ]] = col_num ;
vis [ stack[ top-- ] ] = false ;
}
top -- ;
}
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=;i<m;i++)
{
int x,y,z;
char s[];
scanf("%d%d%d%s",&x,&y,&z,s);
if(s=="AND")
{if(z==)addedge(*x+,*x),addedge(*y+,*y);
else addedge(*x,*y+),addedge(*y,*x+);
}else if(s=="OR")
{if(z==)addedge(*x+,*y),addedge(*y+,*x);
else addedge(*x,*x+),addedge(*y,*y+);
}else{
if(z==)addedge(*x,*y+),addedge(*y+,*x),addedge(*y,*x+),addedge(*x+,*y);
else addedge(*x,*y),addedge(*y,*x),addedge(*x+,*y+),addedge(*y+,*x+);
}
}

 for(int i=;i<\*n;i++)  
     if(!dfn\[i\])Tarjan(i);  
 for(int i=;i<n;i++)  
 if(color\[\*i\]==color\[\*i+\]){printf("NO\\n");return ;}  
 printf("YES\\n");  
 return ;  

}