天梯赛L2题解（029-032）

L2-029 特立独行的幸福

题面看了犯恶心，没见过这么读起来这么难受的题面，遍历每个数的时候，对每个数操作，如果路径里有的答案就不要了(因为显然这个数是个依附别的数的数)，然后注意一下当前这个数要不是依附别的数的数才能算进答案里

#include <bits/stdc++.h>
using namespace std;
#define int long long

int isp(int x) {
    if (x == 1)
        return 1;
    for (int i = 2; i * i <= x; ++i) {
        if (x % i == 0)
            return 1;
    }
    return 2;
}

void ylh_() {
    int l, r;
    cin >> l >> r;
    map<int, int> mp;
    set<int> ex;
    for (int i = l; i <= r; ++i) {
        int cur = i;
        int cnt = 0;
        set<int> st;
        st.insert(cur);
        while (1) {
            int sum = 0;
            while (cur) {
                sum += (cur % 10) * (cur % 10);
                cur /= 10;
            }
            cur = sum;
            if (mp.count(cur)) {
                mp.erase(cur);
            }
            if (cur == 1)
                break;
            if (st.count(cur))
                break;
            st.insert(cur);
            ex.insert(cur);
        }
        if (cur == 1 && !ex.count(i)) {
            mp[i] = st.size() * (isp(i));
            ex.insert(i);
        }
    }
    if (mp.size() == 0) {
        cout << "SAD";
        return;
    }
    for (auto [a, b] : mp) {
        cout << a << ' ' << b << '\n';
    }
}

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--) {
        ylh_();
    }
}

L2-030 冰岛人

注意一种情况就是，可能两个人的公共祖先存在，但是只在一个人的五代以内
因为没给数据范围写了一发暴力发现可以过

#include <bits/stdc++.h>
using namespace std;
#define int long long
using PII = pair<int, int>;

void ylh_() {
    int n;
    cin >> n;
    map<string, int> sex;
    map<string, string> fa;
    for (int i = 1; i <= n; ++i) {
        string x, y;
        cin >> x >> y;
        char ed = *(y.end() - 1);
        if (ed == 'm') {
            sex[x] = 1;
            fa[x] = "-1";
        } else if (ed == 'f') {
            sex[x] = 0;
            fa[x] = "-1";
        } else if (ed == 'n') {
            sex[x] = 1;
            for (int j = 1; j <= 4; ++j)
                y.pop_back();
            fa[x] = y;
        } else {
            sex[x] = 0;
            for (int j = 1; j <= 7; ++j) {
                y.pop_back();
            }
            fa[x] = y;
        }
    }
    auto check = [&](string x, string y) -> bool {
        vector<string> fx, fy;
        string cur = x;
        fx.push_back(x);
        for (int i = 0; i < 3; i++) {
            if (fa[cur] != "-1") {
                cur = fa[cur];
                fx.push_back(cur);
            } else {
                break;
            }
        }
        cur = y;
        fy.push_back(y);
        for (int i = 0; i < 3; i++) {
            if (fa[cur] != "-1") {
                cur = fa[cur];
                fy.push_back(cur);
            } else {
                break;
            }
        }
        for (auto s1 : fx) {
            for (string j = y; j != "-1"; j = fa[j]) {
                if (s1 == j) {
                    return false;
                }
            }
        }
        for (auto s2 : fy) {
            for (string j = x; j != "-1"; j = fa[j]) {
                if (s2 == j) {
                    return false;
                }
            }
        }
        return true;
    };
    int q;
    cin >> q;
    while (q--) {
        string x, y, z;
        cin >> x >> z >> y >> z;
        if (!sex.count(x) || !sex.count(y)) {
            cout << "NA\n";
            continue;
        }
        if (sex[x] == sex[y]) {
            cout << "Whatever\n";
            continue;
        }
        if (check(x, y)) {
            cout << "Yes\n";
        } else {
            cout << "No\n";
        }
    }
}

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--) {
        ylh_();
    }
}

L2-031 深入虎穴

注意一号点不是入口，找到入口的方式就是找到入度为0的点就行了

#include <bits/stdc++.h>
using namespace std;
#define int long long
using PII = pair<int, int>;

void ylh_() {
    int n;
    cin >> n;
    int st = 0;
    vector<int> d(n + 1);
    vector<vector<int>> g(n + 1);
    for (int i = 1; i <= n; ++i) {
        int m;
        cin >> m;
        for (int j = 1; j <= m; ++j) {
            int v;
            cin >> v;
            g[i].push_back(v);
            ++d[v];
        }
    }
    for (int i = 1; i <= n; ++i) {
        if (d[i] == 0) {
            st = i;
        }
    }
    int ans = 1;
    int mxdep = 1;
    auto dfs = [&](auto&& dfs, int u, int dep) -> void {
        if (dep > mxdep) {
            mxdep = dep;
            ans = u;
        }
        for (auto v : g[u]) {
            dfs(dfs, v, dep + 1);
        }
    };
    dfs(dfs, st, 1);
    cout << ans;
}

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--) {
        ylh_();
    }
}

L2-032 彩虹瓶

用个栈表示货架就行了

#include <bits/stdc++.h>
using namespace std;
#define int long long
using PII = pair<int, int>;

void ylh_() {
    int n, m, k;
    cin >> n >> m >> k;
    stack<int> st;
    for (int i = 1; i <= k; ++ i) {
        bool ok = 1;
        int cur = 1;
        for (int j = 1; j <= n; ++ j) {
            int x;
            cin >> x;
            if (x == cur) {
                ++ cur;
                while (!st.empty() && st.top() == cur) {
                    ++ cur;
                    st.pop();
                }
            } else {
                if (st.size() < m) {
                    st.push(x);
                } else {
                    ok = 0;
                }
            }
        }
        if (st.size()) {
            ok = 0;
        }
        if (ok) {
            cout << "YES\n";
        } else {
            cout << "NO\n";
        }
        while (st.size()) st.pop();
    }
}

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--) {
        ylh_();
    }
}