Poj3678：Katu Puzzle

大概题意

有\(n\)个数，可以为\(0/1\)，给\(m\)个条件，表示某两个数经过\(or, and, xor\)后的数是多少

判断是否有解

Sol

\(2-SAT\)判定

建图

# include <iostream>
# include <stdio.h>
# include <stdlib.h>
# include <string.h>
# include <math.h>
# include <algorithm>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2005);
const int __(4e6 + 5);

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, m, first[_], cnt, num;
int S[_], vis[_], dfn[_], low[_], Index, col[_];
struct Edge{
	int to, next;
} edge[__];

IL void Add(RG int u, RG int v){
	edge[cnt] = (Edge){v, first[u]}; first[u] = cnt++;
}

IL void Tarjan(RG int u){
	vis[u] = 1, dfn[u] = low[u] = ++Index, S[++S[0]] = u;
	for(RG int e = first[u]; e != -1; e = edge[e].next){
		RG int v = edge[e].to;
		if(!dfn[v]) Tarjan(v), low[u] = min(low[u], low[v]);
		else if(vis[v]) low[u] = min(low[u], dfn[v]);
	}
	if(dfn[u] != low[u]) return;
	RG int v = S[S[0]--]; col[v] = ++num, vis[v] = 0;
	while(v != u) v = S[S[0]--], col[v] = num, vis[v] = 0;
}

int main(RG int argc, RG char* argv[]){
	Fill(first, -1), n = Input(), m = Input();
	for(RG int i = 1; i <= m; ++i){
		RG int u = Input() + 1, v = Input() + 1, w = Input();
		RG char op; scanf(" %c", &op);
		if(op == 'A'){
			if(w) Add(u, v), Add(v, u), Add(u + n, u), Add(v + n, v);
			else Add(u, v + n), Add(v, u + n);
		}
		else if(op == 'O'){
			if(w) Add(u + n, v), Add(v + n, u);
			else Add(u, u + n), Add(v, v + n), Add(u + n, v + n), Add(v + n, u + n);
		}
		else{
			if(w) Add(u, v + n), Add(v, u + n), Add(u + n, v), Add(v + n, u);
			else Add(u, v), Add(v, u), Add(u + n, v + n), Add(v + n, u + n);
		}
	}
	for(RG int i = 1, tmp = n << 1; i <= tmp; ++i)
		if(!dfn[i]) Tarjan(i);
	for(RG int i = 1; i <= n; ++i)
		if(col[i] == col[i + n]) return puts("NO"), 0;
	return puts("YES"), 0;
}