2L 水一帖，应该已经结束了。

1. 并查集

   #include <bits/stdc++.h>
   using namespace std;
   const int N = 1e5 + 9;
   struct UnionFind {
    int id[N];
    int sz[N];
    int cnt;
    int n;
    UnionFind(int n) : n(n) {
        for (int i = 0; i < n; i++) {
            id[i] = i;
            sz[i] = 1;
        }
        cnt = n;
    }
    int find(int x) {
        if (x == id[x]) {
            return id[x];
        }
        return id[x] = find(id[x]);
    }
    void unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return;
        }
        if (sz[x] > sz[y]) {
            id[y] = x;
            sz[x] += sz[y];
        } else {
            id[x] = y;
            sz[y] += sz[x];
        }
    }
    int getCnt() {
        int ans = 0;
        for (int i = 0; i < n; i++) {
            if (id[i] == i) {
                ans++;
            }
        }
        return ans;
    }
   };
   int main() {
     int n;
     scanf("%d", &n);
     UnionFind* unionFind = new UnionFind(n);
     for (int i = 0; i < n; i++) {
         for (;;) {
             int x; 
             scanf("%d", &x);
             if (x == 0) {
                 break;
             }
             unionFind->unite(i, x-1);
         }
    }
    printf("%d\n", (unionFind->getCnt()));
   }
   

2. 不会，骗了 10%
3. 字符串最小表示法

   #include <bits/stdc++.h>
   using namespace std;
   string mini(string s) {
    string ss(s);
    ss = ss + ss;
    bool flag = false;
    int i = 0, j = 1, k = 0, l = ss.size() / 2, p = 0;
    while (i < l && j < l) {
        k = 0;
        while (ss[i + k] == ss[j + k] && k < l) k++;
        if (k == l) {
            p = i;
            flag = true;
            break;
        }
        if (ss[i + k] > ss[j + k])
            if (i + k + 1 > j) i = i + k + 1; 
            else i = j + 1;
        else if (j + k + 1 > i) j = j + k + 1;
            else j = i + 1;
    }
    if (!flag)
        if (i < j) p = i; 
        else p = j;
    return ss.substr(p, l);
   }
   const int N = 1e5 + 9;
   string a[N];
   string b[N];
   bool solve() {
    int n;
    scanf("%d\n", &n);
    for (int i = 0; i < n; i++) {
        getline(cin, a[i]);
        int len = a[i].size();
        b[i].clear();
        for (int j = len - 1; j >= 0; j--) {
            char c = a[i][j];
            if (c == ' ') c = '*';
            b[i].push_back(c);
        }
        a[i] = mini(a[i]);
        b[i] = mini(b[i]);
    }
    unordered_set<string> st1;
    for (int i = 0; i < n; i++) {
        if (st1.count(a[i])) {
            return true;
        }
        st1.insert(a[i]);
    }
    unordered_set<string> st2;
    for (int i = 0; i < n; i++) {
        if (st2.count(b[i])) {
            return true;
        }
        st2.insert(b[i]);
    }
    for (int i = 0; i < n; i++) {
        if (st1.count(b[i]) && a[i] != b[i]) {
            return true;
        }
    }
    return false;
   }
   int main() {
    int t;
    scanf("%d\n", &t);
    while (t--) {
        if (solve()) {
            puts("Yeah");
        } else {
            puts("Sad");
        }
    }
   }
   

4. 不会，用t = 1时候的用最长不降子序列骗了 20 %
5. 根本不会。

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