【HDU】4421 Bit Magic 2-sat

传送门：【HDU】4421 Bit Magic

题目分析：将一个数拆成31位，分别用2-sat判断是否符合即可。

代码如下：

#include <map>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#pragma comment ( linker , "/STACK:1024000000" )
#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define rec( i , A , o ) for ( int i = A[o] ; i != o ; i = A[i] )
#define clr( a , x ) memset ( a , x , sizeof a )

const int MAXN = 505 ;
const int MAXE = 1000005 ;

struct Edge {
	int v , n ;
	Edge () {}
	Edge ( int v , int n ) : v ( v ) , n ( n ) {}
} ;

Edge E[MAXE] ;
int H[MAXN << 1] , cntE ;
int scc[MAXN << 1] , scc_cnt ;
int dfn[MAXN << 1] , low[MAXN << 1] , dfs_clock ;
int S[MAXN << 1] , top ;
int b[MAXN][MAXN] ;
int n ;

void clear () {
	top = 0 ;
	cntE = 0 ;
	scc_cnt = 0 ;
	dfs_clock = 0 ;
	clr ( H , -1 ) ;
	clr ( scc , 0 ) ;
	clr ( dfn , 0 ) ;
}

void addedge ( int u , int v ) {
	E[cntE] = Edge ( v , H[u] ) ;
	H[u] = cntE ++ ;
}

void tarjan ( int u ) {
	dfn[u] = low[u] = ++ dfs_clock ;
	S[top ++] = u ;
	for ( int i = H[u] ; ~i ; i = E[i].n ) {
		int v = E[i].v ;
		if ( !dfn[v] ) {
			tarjan ( v ) ;
			low[u] = min ( low[u] , low[v] ) ;
		} else if ( !scc[v] ) low[u] = min ( low[u] , dfn[v] ) ;
	}
	if ( low[u] == dfn[u] ) {
		++ scc_cnt ;
		do {
			scc[S[-- top]] = scc_cnt ;
		} while ( u != S[top] ) ;
	}
}

bool check () {
	rep ( i , 0 , n ) if ( scc[i << 1] == scc[i << 1 | 1] ) return 0 ;
	return 1 ;
}

void scanf ( int& x , char c = 0 ) {
	while ( ( c = getchar () ) < '0' || c > '9' ) ;
	x = c - '0' ;
	while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ;
}

void solve () {
	rep ( i , 0 , n ) rep ( j , 0 , n ) scanf ( b[i][j] ) ;
	rep ( i , 0 , 31 ) {
		clear () ;
		rep ( j , 0 , n ) {
			rep ( k , 0 , n ) {
				if ( j == k ) continue ;
				if ( j % 2 == 1 && k % 2 == 1 ) {
					if ( ( b[j][k] >> i ) & 1 ) {
						addedge ( j << 1 | 1 , k << 1 ) ;
						addedge ( k << 1 | 1 , j << 1 ) ;
					} else {
						addedge ( j << 1 | 1 , k << 1 | 1 ) ;
						addedge ( k << 1 | 1 , j << 1 | 1 ) ;
					}
				} else if ( j % 2 == 0 && k % 2 == 0 ) {
					if ( ( b[j][k] >> i ) & 1 ) {
						addedge ( j << 1 , k << 1 ) ;
						addedge ( k << 1 , j << 1 ) ;
					} else {
						addedge ( j << 1 , k << 1 | 1 ) ;
						addedge ( k << 1 , j << 1 | 1 ) ;
					}
				} else {
					if ( ( b[j][k] >> i ) & 1 ) {
						addedge ( j << 1 , k << 1 | 1 ) ;
						addedge ( k << 1 , j << 1 | 1 ) ;
						addedge ( j << 1 | 1 , k << 1 ) ;
						addedge ( k << 1 | 1 , j << 1 ) ;
					} else {
						addedge ( j << 1 , k << 1 ) ;
						addedge ( k << 1 , j << 1 ) ;
						addedge ( j << 1 | 1 , k << 1 | 1 ) ;
						addedge ( k << 1 | 1 , j << 1 | 1 ) ;
					}
				}
			}
		}
		rep ( u , 0 , n ) if ( !dfn[u] ) tarjan ( u ) ;
		if ( !check () ) {
			printf ( "NO\n" ) ;
			return ;
		}
	}
	printf ( "YES\n" ) ;
}

int main () {
	while ( ~scanf ( "%d" , &n ) ) solve () ;
	return 0 ;
}