Solution

给你一些父子关系，判断这个些点构成的集合是否是一棵树，内存给的很小。
使用并查集维护，如果集合内部合并说明形成了环，不是一棵树，多个根节点，也不是一棵树。
最后注意特判一下空树的情况就可以过了。

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <iostream>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); return; }    char F[40]; ll tmp = x > 0 ? x : -x;    if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]);    if (op)    putchar(op); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;

const int N = 100 + 7;

int fa[N], vis[N], tmp[N];
int find(int x) {
    return fa[x] == x ? x : fa[x] = find(fa[x]);
}

int main() {
    js;
    int ca = 0, a, b;
    while (cin >> a >> b) {
        if (a == -1 and b == -1)
            break;
        if (a == 0) { //特判空树
            cout << "Case " << ++ca << " is a tree." << endl;
            continue;
        }
        ms(vis, 0), ms(tmp, 0);
        for (int i = 0; i < N; ++i)
            fa[i] = i;
        int flag = 1;
        fa[b] = a;
        tmp[a] = tmp[b] = vis[b] = 1;
        while (cin >> a >> b) {
            if (a == 0 and b == 0)
                break;
            tmp[a] = tmp[b] = 1;
            int A = find(a), B = find(b);
            if (A != B)
                fa[B] = A;
            else
                flag = 0;
            if (++vis[b] > 1)
                flag = 0;
        }
        int cnt = 0;
        for (int i = 1; i < N; ++i)
            if (tmp[i] and vis[i] == 0)
                ++cnt; //找独立的根
        if (cnt == 1 and flag == 1)
            cout << "Case " << ++ca << " is a tree." << endl;
        else
            cout << "Case " << ++ca << " is not a tree." << endl;
    }
    return 0;
}