天梯赛L2题解（049-052）

L2-049 鱼与熊掌

set搞一下就行了

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

void ylh_() {
    int n, m;
    cin >> n >> m;
    vector<set<int>> st(n + 1);
    for (int i = 1; i <= n; ++i) {
        int k;
        cin >> k;
        for (int j = 1; j <= k; ++j) {
            int x;
            cin >> x;
            st[i].insert(x);
        }
    }
    int q;
    cin >> q;
    while (q--) {
        int x, y;
        cin >> x >> y;
        int ans = 0;
        for (int i = 1; i <= n; ++i) {
            if (st[i].count(x) && st[i].count(y)) {
                ++ans;
            }
        }
        cout << ans << '\n';
    }
}

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

L2-050 懂蛇语

注意空格分隔不止一个空格，所以要用getline读入

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

void ylh_() {
    int n;
    cin >> n;
    map<string, vector<string>> mp;
    for (int i = 1; i <= n; ++i) {
        string s = "", t = "";
        if (i == 1) {
            getline(cin, s);
        }
        getline(cin, s);
        for (int i = 0; i < s.length(); ++i) {
            if (s[i] >= 'a' && s[i] <= 'z' && (i == 0 || s[i - 1] == ' ')) {
                t += s[i];
            }
        }
        mp[t].push_back(s);
    }
    for (auto& [x, vec] : mp) {
        sort(vec.begin(), vec.end());
    }
    int m;
    cin >> m;
    for (int i = 1; i <= m; ++i) {
        string s = "", t = "";
        if (i == 1) {
            getline(cin, s);
        }
        getline(cin, s);
        for (int i = 0; i < s.length(); ++i) {
            if (s[i] >= 'a' && s[i] <= 'z' && (i == 0 || s[i - 1] == ' ')) {
                t += s[i];
            }
        }
        if (!mp[t].size()) {
            cout << s << '\n';
        }
        for (int i = 0; i < mp[t].size(); ++i) {
            cout << mp[t][i] << "|\n"[i == mp[t].size() - 1];
        }
    }
}

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

L2-051 满树的遍历

其实我不太懂这种非二叉树的前序遍历是个什么玩意，于是按顺序遍历试了一下发现过了

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

const int MOD = 998244353;
using PII = pair<int, int>;

void ylh_() {
    int n;
    cin >> n;
    vector<vector<int>> g(n + 1);
    int rt = -1;
    vector<int> degree(n + 1); // 度数的出现次数
    for (int i = 1; i <= n; ++i) {
        int x;
        cin >> x;
        if (x == 0)
            rt = i;
        else {
            g[x].push_back(i);
        }
    }
    vector<int> ans;
    auto dfs = [&](auto&& dfs, int u) -> void {
        int deg = 0;
        ans.push_back(u);
        for (auto v : g[u]) {
            dfs(dfs, v);
            ++deg;
        }
        degree[deg]++;
    };
    dfs(dfs, rt);
    int cnt = 0;
    int Max = -1;
    for (int i = 0; i <= n; ++i) {
        if (degree[i] != 0) {
            ++cnt;
            Max = i;
        }
    }
    if (cnt >= 3) {
        cout << Max << ' ' << "no" << '\n';
    } else {
        cout << Max << ' ' << "yes" << '\n';
    }
    for (int i = 0; i < n; ++i) {
        cout << ans[i] << " \n"[i == n - 1];
    }
}

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

L2-052 吉利矩阵

其实就是爆搜剪枝，我想了一下数据小就没仔细思考了，每次剪一点点剪了三次小点就过了
可以专门用数组记录行列和，然后超出就剪枝
实际上可以根据每行或者每列的前n -1个数确定最后一个数，可以缩小指数
甚至可以得出答案后打表也是可以的

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

using PII = pair<int, int>;

void ylh_() {
    int l, n;
    cin >> l >> n;
    vector<vector<int>> a(n + 1, vector<int>(n + 1));
    int ans = 0;
    auto dfs = [&](auto&& dfs, int x, int y) -> void {
        if (x == n + 1) {
            for (int j = 1; j <= n; j++) {
                int s = 0;
                for (int i = 1; i <= n; i++)
                    s += a[i][j];
                if (s != l)
                    return;
            }
            ans++;
            return;
        }
        if (y == n + 1) {
            int s = 0;
            for (int i = 1; i <= n; i++)
                s += a[x][i];
            if (s != l)
                return;
            dfs(dfs, x + 1, 1);
            return;
        }
        int s = 0;
        for (int i = 1; i < y; i++)
            s += a[x][i];
        if (s > l)
            return;
        s = 0;
        for (int i = 1; i < x; ++i) {
            s += a[i][y];
        }
        if (s > l)
            return;
        for (int v = 0; v <= l; v++) {
            if (s + v > l)
                continue;
            a[x][y] = v;
            if (x == n) {
                int t = 0;
                for (int i = 1; i <= n; i++)
                    t += a[i][y];
                if (t != l)
                    continue;
            }
            dfs(dfs, x, y + 1);
        }
    };
    dfs(dfs, 1, 1);
    cout << ans << '\n';
}

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