// 题目来源：POJ 3678// 题目大意：有很多点可取值0或1，它们之间有很多边为0或1，规定每条边由端点（xor，and，or）得到，求是否可行// 解决方法：每个点的取值只有两种情况，非常符合2-sat，用强连通分量判断即可// 特别注意：可通过在拆成的点对中搭边来保证最后的结果落在某个点上#include <cstdio>#include <string>#define o 1002#define _ 1000002void link( int, int );void tarjan( int );int next[ _*4 ], g[ _*4 ], h[ o*4 ], code[ o*4 ], dfn[ o*4 ], low[ o*4 ], stack[ o*4 ];int n, m, index, cnt, t, top;bool ins[ o*4 ];int main( ){ freopen( "3678.in", "r", stdin ); freopen( "3678.out", "w", stdout ); scanf( "%d%d", &n, &m ); int aa, bb, cc; char st[ 10 ]; for( int i = 1; i <= m; i++ ) { scanf( "%d%d%d%s", &aa, &bb, &cc, st ); if( strcmp( st, "AND" ) == 0 ) if( cc == 0 ) { link( aa+n, bb ); link( bb+n, aa ); } else { link( aa+n, bb+n ); link( bb+n, aa+n ); link( aa, aa+n ); link( bb, bb+n ); } if( strcmp( st, "OR" ) == 0 ) if( cc == 0 ) { link( aa, bb ); link( bb, aa ); link( aa+n, aa ); link( bb+n, bb ); } else { link( aa, bb+n ); link( bb, aa+n ); } if( strcmp( st, "XOR" ) == 0 ) if( cc == 0 ) { link( aa, bb ); link( bb, aa ); link( aa+n, bb+n ); link( bb+n, aa+n ); } else { link( aa, bb+n ); link( bb, aa+n ); link( aa+n, bb ); link( bb+n, aa ); } } for( int i = 0; i < 2*n; i++ ) if( !dfn[ i ] ) tarjan( i ); for( int i = 0; i < n; i++ ) if( code[ i ] == code[ i + n ] ) { printf( "NO" ); return 0; } printf( "YES" ); return 0;}void link( int aa, int bb ){ next[ ++t ] = h[ aa ]; h[ aa ] = t; g[ t ] = bb;}void tarjan( int i ){ int j; dfn[ i ] = low[ i ] = ++index; stack[ ++top ] = i; ins[ i ] = 1; for( int k = h[ i ]; k; k = next[ k ] ) { j = g[ k ]; if( !dfn[ j ] ) { tarjan( j ); if( low[ j ] < low[ i ] ) low[ i ] = low[ j ]; } else if( ins[ j ] && dfn[ j ] < low[ i ] ) low[ i ] = dfn[ j ]; } if( dfn[ i ] == low[ i ] ) { cnt++; do { j = stack[ top-- ]; code[ j ] = cnt; ins[ j ] = 0; } while( i != j ); }}
