直接给我掉！

有心了

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 510, M = 2010;
int n, m;
int p[N], d[N];
int a[M], b[M], op[M]; // 存储输入数据

int find(int x) {
    if (p[x] != x) {
        int root = find(p[x]);
        d[x] = (d[x] + d[p[x]]) % 3;
        p[x] = root;
    }
    return p[x];
}

int main() {
    while (cin >> n >> m) {
        for (int i = 0; i < m; i++) {
            char c;
            cin >> a[i] >> c >> b[i];
            if (c == '=') op[i] = 0;
            else if (c == '<') op[i] = 1; // a < b 表示 a 输给 b
            else op[i] = 2; // a > b 表示 a 赢 b
        }

        int judge = -1, max_line = 0;
        int cnt = 0; // 可能为裁判的人数

        // 枚举假设裁判
        for (int i = 0; i < n; i++) {
            // 初始化并查集
            for (int j = 0; j < n; j++) p[j] = j;
            memset(d, 0, sizeof d);

            int line = 0; // 记录发生矛盾的轮数（如果发生）
            for (int k = 0; k < m; k++) {
                int x = a[k], y = b[k];
                if (x == i || y == i) continue; // 跳过涉及裁判的轮次

                int fx = find(x), fy = find(y);
                if (fx == fy) {
                    // 同根，检查关系是否矛盾
                    if ((d[x] - d[y] + 3) % 3 != op[k]) {
                        line = k + 1; // 记录矛盾轮数（从1开始计）
                        break;
                    }
                } else {
                    // 合并
                    p[fx] = fy;
                    d[fx] = (d[y] - d[x] + op[k] + 3) % 3;
                }
            }
            if (line == 0) {
                // 假设 i 是裁判没有矛盾
                cnt++;
                judge = i;
            } else {
                // 记录排除其他候选人的最大轮数
                max_line = max(max_line, line);
            }
        }

        // 输出结果
        if (cnt == 0) {
            cout << "Impossible" << endl;
        } else if (cnt > 1) {
            cout << "Can not determine" << endl;
        } else {
            cout << "Player " << judge << " can be determined to be the judge after " << max_line << " lines" << endl;
        }
    }
    return 0;
}

g'g'f'd'g'd