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;
}