题目链接

思路

既然\(x\)的数量那么小，我们就可以先把每个\(x\)搜索一遍。

枚举x的时候不需要把\(a,b,c\)全枚举一遍，只要枚举其中的两个就可以枚举到当前位置选任何车的情况。

然后就变成了只有\('a','b','c'\)的序列。寻找满足题目要求的方案。

\(2-sat\)模型。

连边的时候注意一些技巧，否则\(if\)写到自闭。。

在\(UOJ\)上会被卡掉\(3\)分。实在懒得去卡常了233

代码

/*
* @Author: wxyww
* @Date:   2019-04-29 09:08:01
* @Last Modified time: 2019-04-30 14:32:38
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 100010;
ll read() {
    ll x=0,f=1;char c=getchar();
    while(c<'0'||c>'9') {
        if(c=='-') f=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9') {
        x=x*10+c-'0';
        c=getchar();
    }
    return x*f;
}
vector<int>e[N];
char a[N],tmp[N];
int n,D,m;
int opp[N];
#define add(x,y) e[x].push_back(y)
int val[N];

int tot,dfn[N],low[N],vis[N],sta[N],coljs,col[N],top;

void tarjan(int u) {
    int k = e[u].size();
    dfn[u] = low[u] = ++tot;
    sta[++top] = u;vis[u] = 1;
    for(int i = 0;i < k;++i) {
        int v = e[u][i];
        if(!dfn[v]) {
            tarjan(v);low[u] = min(low[u],low[v]);
        }
        else if(vis[v]) low[u] = min(low[u],low[v]);
    }
    if(dfn[u] == low[u]) {
        ++coljs;
        do {
            int x = sta[top--];
            col[x] = coljs;
            vis[x] = 0;
        }while(sta[top + 1] != u);
    }
}

int t1[N],t2[N];
char h1[N],h2[N];

int get_u(int x,char y) {
    if(a[x] == 'a') return y == 'B' ? x : x + n;
    return y == 'A' ? x : x + n;
}

#define ote(x) x > n ? x - n : x + n

void solve() {
    memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));
    memset(col,0,sizeof(col));
    coljs = 0,tot = 0;top = 0;
    for(int i = 1;i <= n + n;++i) e[i].clear();
    for(int i = 1;i <= m;++i) {
        if(a[t1[i]] == h1[i] - 'A' + 'a') continue;
        int u = get_u(t1[i],h1[i]);
        if(a[t2[i]] == h2[i] - 'A' + 'a') {
            add(u,ote(u));
            continue;
        }
        int v = get_u(t2[i],h2[i]);
        add(u,v);add(ote(v),ote(u));
    }
    for(int i = 1;i <= n + n;++i) if(!dfn[i]) tarjan(i);

    for(int i = 1;i <= n;++i) if(col[i] == col[i + n]) return;

    for(int i = 1;i <= n;++i) {
        if(col[i] < col[i + n]) {
            if(a[i] == 'a') putchar('B');
            else putchar('A');
        }
        else {
            if(a[i] == 'c') putchar('B');
            else putchar('C');
        }
    }
    exit(0);
}

void dfs(int pos) {
    if(pos == n + 1) {
        solve();
        return;
    }
    if(tmp[pos] != 'x') dfs(pos + 1);
    else {
        a[pos] = 'a';
        dfs(pos + 1);
        a[pos] = 'b';
        dfs(pos + 1);
    }
}

int main() {
    // freopen("2305/game3.in","r",stdin);
    n = read(),D = read();
    scanf("%s",tmp + 1);
    memcpy(a + 1,tmp + 1,n);
    m = read();
    for(int i = 1;i <= m;++i) {
        t1[i] = read();h1[i] = getchar();while(h1[i] != 'A' && h1[i] != 'B' && h1[i] != 'C') h1[i] = getchar();
        t2[i] = read();h2[i] = getchar();while(h2[i] != 'A' && h2[i] != 'B' && h2[i] != 'C') h2[i] = getchar();
    }

    dfs(1);
    puts("-1");
    return 0;
}

原文地址：https://www.cnblogs.com/wxyww/p/loj2305.html