POJ 3678 2-SAT

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<cstdio>
 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 ;
 }