Sol
然而我的代码在\(UOJ\)上被\(hack\)\(\ TLE\)了
但思路是没问题的\(TAT\)
如果没有\(x\)，就是个\(2-SAT\)
我们爆搜\(x\)的地图是\(a\)还是\(b\)就好了
不用枚举它是\(c\)，枚举\(a\)，\(b\)就能保证正好选\(ABC\)三种车
我也不知道我的输出方案哪里学的
拓扑排序+染色\(QAQ\)
如果觉得自己的代码优秀就去\(UOJ\)上交吧(逃
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e5 + 5);
typedef int Arr[_];
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;
}
Arr dfn, low, S, vis, first, head, col, i1, i2, d, oth, cho;
int n, m, D, cnt, pos[10], idx, num;
struct Edge{
int to, next;
} edge[_ << 2], dag[_ << 2];
char s[_], h1[_], h2[_], tp1[200], tp2[200];
queue <int> Q;
IL void Add(RG int u, RG int v){
edge[cnt] = (Edge){v, first[u]}, first[u] = cnt++;
}
IL void Add_DAG(RG int u, RG int v){
dag[cnt] = (Edge){v, head[u]}, head[u] = cnt++;
}
IL void Tarjan(RG int u){
dfn[u] = low[u] = ++idx, vis[S[++S[0]] = u] = 1;
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(low[u] == dfn[u]){
RG int v = S[S[0]--]; vis[v] = 0, col[v] = ++num;
while(v != u) vis[v = S[S[0]--]] = 0, col[v] = num;
}
}
IL void Del(RG int u){
cho[u] = 0;
for(RG int e = head[u]; e != -1; e = dag[e].next)
if(cho[u] == -1) Del(dag[e].to);
}
IL void Calc(){
RG int tot = n + n; num = idx = cnt = 0;
for(RG int i = 1; i <= tot; ++i)
cho[i] = head[i] = first[i] = -1, d[i] = vis[i] = oth[i] = dfn[i] = low[i] = col[i] = 0;
for(RG int i = 1; i <= m; ++i){
if(s[i1[i]] == h1[i]) continue;
if(i1[i] == i2[i] && h1[i] == h2[i]) continue;
RG int p1 = i1[i], p2 = i1[i] + n, p3 = i2[i], p4 = i2[i] + n;
if(tp1[s[i1[i]]] != h1[i]) swap(p1, p2);
if(tp1[s[i2[i]]] != h2[i]) swap(p3, p4);
if(s[i2[i]] == h2[i]) Add(p1, p2);
else Add(p1, p3), Add(p4, p2);
}
for(RG int i = 1; i <= tot; ++i) if(!dfn[i]) Tarjan(i);
for(RG int i = 1; i <= n; ++i){
if(col[i] == col[i + n]) return;
oth[col[i]] = col[i + n], oth[col[i + n]] = col[i];
}
cnt = 0;
for(RG int i = 1; i <= tot; ++i)
for(RG int e = first[i]; e != -1; e = edge[e].next)
if(col[edge[e].to] != col[i]) Add_DAG(col[edge[e].to], col[i]), ++d[col[i]];
for(RG int i = 1; i <= num; ++i) if(!d[i]) Q.push(i);
while(!Q.empty()){
RG int u = Q.front(); Q.pop();
if(cho[u] != -1) continue;
cho[u] = 1, Del(oth[u]);
for(RG int e = head[u]; e != -1; e = dag[e].next)
if(!--d[dag[e].to]) Q.push(dag[e].to);
}
for(RG int i = 1; i <= n; ++i)
(cho[col[i]] == 1) ? putchar(tp1[s[i]] - 'a' + 'A') : putchar(tp2[s[i]] - 'a' + 'A');
puts(""); exit(0);
}
IL void Dfs(RG int x){
if(x > D){
Calc();
return;
}
s[pos[x]] = 'a', Dfs(x + 1), s[pos[x]] = 'b', Dfs(x + 1);
}
int main(RG int argc, RG char *argv[]){
tp1['a'] = 'b', tp1['b'] = 'c', tp1['c'] = 'a';
tp2['a'] = 'c', tp2['b'] = 'a', tp2['c'] = 'b';
n = Input(), D = Input(), scanf(" %s", s + 1), m = Input();
for(RG int i = 1; i <= m; ++i)
i1[i] = Input(), h1[i] = getchar() - 'A' + 'a', i2[i] = Input(), h2[i] = getchar() - 'A' + 'a';
for(RG int i = 1, t = 0; i <= n; ++i) if(s[i] == 'x') pos[++t] = i;
Dfs(1);
puts("-1");
return 0;
}