容易想到拓扑排序。
然后做完了，记一个 flag 表示是否有同时入队的情况。

#include <bits/stdc++.h>
using namespace std;
const int N = 30, M = 615;
int n, m, h[N], e[M], ne[M], idx = 0;
void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

int d[N], in[N], ans[N];
int topsort() {
    for (int i = 1; i <= n; i++) in[i] = d[i];
    queue<int> q;
    bool flag = 1;
    int cnt = 0, tot = 0;
    for (int i = 1; i <= n; i++)
        if (!in[i]) q.push(i), cnt++;
    flag &= (cnt == 1);
    while (q.size()) {
        int u = q.front(); q.pop();
        ans[++tot] = u;
        cnt = 0;
        for (int i = h[u]; ~i; i = ne[i]) {
            int v = e[i];
            --in[v];
            if (!in[v]) cnt++, q.push(v);
        }
        flag &= (cnt <= 1);
    }
    if (tot != n) return -1;    // 有环
    if (flag == 1) return 1;    // 可以唯一确定
    return 0;
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) h[i] = -1;
    for (int i = 1; i <= m; i++) {
        char ch[5];
        scanf("\n %s", ch + 1);
        int x = ch[1] - 'A' + 1, y = ch[3] - 'A' + 1;
        add(x, y), d[y]++;
        int result = topsort();
        if (result == -1) return printf("Inconsistency found after %d relations.\n", i), 0;
        if (result == 1) {
            printf("Sorted sequence determined after %d relations: ", i);
            for (int j = 1; j <= n; j++) putchar(ans[j] + 'A' - 1);
            putchar('.'), putchar('\n');
            return 0;
        }
    }
    puts("Sorted sequence cannot be determined.");
    return 0;
}