传送门

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第一眼看上去似乎是一个3-SAT问题

然而\(d \leq 8\)给我们的信息就是：暴力枚举

枚举\(x\)型地图变成\(a\)型地图还是\(b\)型地图（实际上不要枚举\(c\)，因为\(ab\)两种地图已经包含了选择\(ABC\)三辆车的情况），对于每一种情况跑2-SAT即可。复杂度\(O(2^d(n+m))\)

还有为什么UOJ的Hack那么强啊QAQ

随机化要么WA EX5要么WA EX8，不随机化TLE EX9

UPD：在Itst的重构之后终于过了UOJ的Hack，只是不知道为什么数组一定要开\(10^5\)

#include<bits/stdc++.h>
//this code is written by Itst
using namespace std;

int read(){
    int a = 0;
    char c = getchar();
    while(!isdigit(c)) c = getchar();
    while(isdigit(c)){
        a = a * 10 + c - 48;
        c = getchar();
    }
    return a;
}

char getc(){
    char c = getchar();
    while(!isupper(c)) c = getchar();
    return c;
}

const int MAXN = 1e5 + 3;
struct Edge{
    int end , upEd;
}Ed[MAXN << 2];
struct que{
    int a , tpa , b , tpb;
}now[MAXN << 1];
char s[MAXN];
int dfn[MAXN << 1] , low[MAXN << 1] , in[MAXN << 1] , head[MAXN << 1];
int N , M , D , ts , cntEd , cntSCC , X[9];
int stk[MAXN << 1] , top;
bool vis[MAXN << 1] , ins[MAXN << 1];

void addEd(int a , int b){
    Ed[++cntEd] = (Edge){b , head[a]};
    head[a] = cntEd;
}

void clear(){
    memset(head , 0 , sizeof(head));
    memset(vis , 0 , sizeof(vis));
    memset(ins , 0 , sizeof(ins));
    memset(in , 0 , sizeof(in));
    cntSCC = cntEd = ts = top = 0;
}

bool pop(int tar){
    ++cntSCC;
    do{
        int t = stk[top];
        in[t] = cntSCC;
        if(in[t > N ? t - N : t + N] == in[t])
            return 0;
    }while(stk[top--] != tar);
    return 1;
}

int ind(int x , int p){
    if(s[x] != 'c') return (p == 2) * N + x;
    return (p == 1) * N + x;
}

bool tarjan(int x){
    vis[x] = ins[x] = 1;
    dfn[x] = low[x] = ++ts;
    stk[++top] = x;
    for(int i = head[x] ; i ; i = Ed[i].upEd){
        if(!vis[Ed[i].end]){
            if(!tarjan(Ed[i].end)) return 0;
        }
        else if(!ins[Ed[i].end]) continue;
        low[x] = min(low[x] , low[Ed[i].end]);
    }
    ins[x] = 0;
    if(dfn[x] == low[x])
        return pop(x);
    return 1;
}

bool work(){
    for(int i = 1 ; i <= M ; ++i){
        bool f1 = now[i].tpa == s[now[i].a] - 'a' , f2 = now[i].tpb == s[now[i].b] - 'a';
        if(f1) continue;
        if(f2){
            int pos = ind(now[i].a , now[i].tpa);
            addEd(pos , now[i].a * 2 + N - pos);
        }
        int posA = ind(now[i].a , now[i].tpa) , posB = ind(now[i].b , now[i].tpb);
        addEd(posA , posB); addEd(now[i].b * 2 + N - posB , now[i].a * 2 + N - posA);
    }
    for(int i = 1 ; i <= 2 * N ; ++i)
        if(!vis[i] && !tarjan(i)) return 0;
    return 1;
}

void output(){
    for(int i = 1 ; i <= N ; ++i)
        if(in[i] > in[i + N])
            putchar(s[i] == 'c' ? 'B' : 'C');
        else putchar(s[i] == 'a' ? 'B' : 'A');
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("in","r",stdin);
    //freopen("out","w",stdout);
#endif
    N = read(); read();
    scanf("%s" , s + 1);
    for(int i = 1 ; i <= N ; ++i)
        if(s[i] == 'x')
            X[++D] = i;
    M = read();
    for(int i = 1 ; i <= M ; ++i){
        now[i].a = read(); now[i].tpa = getc() - 'A';
        now[i].b = read(); now[i].tpb = getc() - 'A';
    }
    for(int i = 0 ; i < 1 << D ; ++i){
        for(int j = 1 ; j <= D ; ++j)
            s[X[j]] = (i >> (j - 1) & 1) + 'a';
        if(work()){
            output();
            return 0;
        }
        clear();
    }
    puts("-1");
    return 0;
}