Solution -「NOI 2017」「洛谷 P3825」游戏

\(\mathscr{Description}\)

  Link.

  给大家看个乐子: link, 懒得概括题意啦.

\(\mathscr{Solution}\)

  对于没有 X 的情况, 显然可以 2-SAT; 对于有 X 的情况, 暴力枚举这辆车属于 \(\{X,A\}\) 还是 \(\{B,C\}\), 继续 2-SAT. 复杂度 \(\mathcal O(2^d(n+m))\).

\(\mathscr{Code}\)

/* Clearink */

#include <cstdio>
#include <cstdlib>

const int MAXN = 5e4, MAXM = 1e5;
int n, cx, m, xpos[10], xid[MAXN + 5];
char s[MAXN + 5];

inline void chkmin ( int& a, const int b ) { b < a && ( a = b, 0 ); }

inline int id ( const int sta, const int i, const int c ) {
	if ( ~xid[i] ) {
		if ( ( sta >> xid[i] ) & 1 ) return c ? -1 : 0;
		if ( !c ) return -1;
		return ( c + 2 ) % 3;
	} else {
		if ( s[i] == c ) return -1;
		return ( c - s[i] + 2 ) % 3;
	}
}

struct Restrict {
	int u, a, v, b; // u use a -> v use b.
	inline void read () {
		char ta, tb;
		scanf ( "%d %c %d %c", &u, &ta, &v, &tb ), a = ta - 'A', b = tb - 'A';
	}
} restr[MAXM + 5];

struct TwoSAT {
	static const int MAXND = MAXN << 1, MAXEG = MAXM << 1;
	int ecnt, head[MAXND + 5], to[MAXEG + 5], nxt[MAXEG + 5];
	int dfc, dfn[MAXND + 5], low[MAXND + 5], top, stk[MAXND + 5];
	int scc, clr[MAXND + 5];
	bool ins[MAXND + 5];

	inline void clear () {
		ecnt = scc = dfc = 0;
		for ( int i = 1; i <= n << 1; ++i ) head[i] = dfn[i] = 0;
	}

	inline void link ( const int s, const int t ) {
		to[++ecnt] = t, nxt[ecnt] = head[s];
		head[s] = ecnt;
	}

	inline void Tarjan ( const int u ) {
		int v;
		dfn[u] = low[u] = ++dfc, ins[stk[++top] = u] = true;
		for ( int i = head[u]; i; i = nxt[i] ) {
			if ( !dfn[v = to[i]] ) Tarjan ( v ), chkmin ( low[u], low[v] );
			else if ( ins[v] ) chkmin ( low[u], dfn[v] );
		}
		if ( dfn[u] == low[u] ) {
			++scc;
			do ins[v = stk[top--]] = false, clr[v] = scc; while ( u ^ v );
		}
	}

	inline bool check ( const int sta ) {
		for ( int i = 1; i <= n << 1; ++i ) if ( !dfn[i] ) Tarjan ( i );
		for ( int i = 1; i <= n; ++i ) if ( clr[i] == clr[i + n] ) return false;
		for ( int i = 1; i <= n; ++i ) {
			bool val = clr[i] > clr[i + n];
			if ( !~xid[i] ) putchar ( ( s[i] + 1 + val ) % 3 + 'A' );
			else if ( !( ( sta >> xid[i] ) & 1 ) ) putchar ( 1 + val + 'A' );
			else putchar ( 'A' );
		}
		return putchar ( '\n' ), exit ( 0 ), true;
	}

	inline void build ( const int sta ) {
		clear ();
		for ( int i = 0; i < cx; ++i ) {
			if ( ( sta >> i ) & 1 ) {
				link ( xpos[i] + n, xpos[i] ); // choose A only.
			}
		}
		for ( int i = 1; i <= m; ++i ) {
			int ia = id ( sta, restr[i].u, restr[i].a );
			int ib = id ( sta, restr[i].v, restr[i].b );
			if ( !~ia ) continue;
			if ( !~ib ) link ( restr[i].u + ia * n, restr[i].u + ( ia ^ 1 ) * n );
			else {
				link ( restr[i].u + ia * n, restr[i].v + ib * n );
				link ( restr[i].v + ( ib ^ 1 ) * n, restr[i].u + ( ia ^ 1 ) * n );
			}
		}
	}
} sat;

int main () {
	scanf ( "%d %d %s", &n, &cx, s + 1 );
	for ( int i = 1, t = 0; i <= n; ++i ) {
		if ( s[i] == 'x' ) {
			xpos[xid[i] = t++] = i;
			s[i] = '$';
		} else s[i] -= 'a', xid[i] = -1;
	}
	scanf ( "%d", &m );
	for ( int i = 1; i <= m; ++i ) restr[i].read ();
	for ( int s = 0; s < 1 << cx; ++s ) {
		sat.build ( s );
		if ( sat.check ( s ) ) return 0;
	}
	puts ( "-1" );
	return 0;
}