Codeforces Round 805 (Div. 3)

原题链接：Codeforces Round 805 (Div. 3)

目录

• A
• B
• C
• D
• E
• F
• G

A

题意：

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思路：

B

题意：

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思路：

C

题意：

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思路：

D

题意：

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思路：

E

题意：判定二部图，并且\(deg \le 2\)

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思路：BFS DSU jiangly判断奇环(仅限于\(deg \le 2\) )

#include <bits/stdc++.h>

using i64 = long long;

void solve() {
    int n;
    std::cin >> n;

    std::vector<std::vector<int>> adj(n);
    for (int i = 0, u, v; i < n; ++i) {
        std::cin >> u >> v;
        --u, --v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    
    std::vector<int> f(n, -1);

    for (int i = 0; i < n; ++i) {
        if (f[i] != -1) {
            continue;
        }
        std::queue<int> q;
        q.push(i);
        f[i] = 0;
        
        while (!q.empty()) {
            auto u = q.front();
            q.pop();

            if (adj[u].size() > 2) {
                std::cout << "NO\n";
                return;
            }
        
            for (auto v : adj[u]) {
                if (f[v] == -1) {
                    f[v] = 1 ^ f[u];
                    q.push(v);
                } else if (f[v] == f[u]) {
                    std::cout << "NO\n";
                    return;
                }
            }
        }
    }
    std::cout << "YES\n";
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int t;
    std::cin >> t;

    while (t--) {
        solve();
    }
}

#include <bits/stdc++.h>

using i64 = long long;

struct DSU {
    std::vector<int> f, siz;
    
    DSU() {}
    DSU(int n) {
        init(n);
    }
    
    void init(int n) {
        f.resize(n);
        std::iota(f.begin(), f.end(), 0);
        siz.assign(n, 1);
    }
    
    int find(int x) {
        while (x != f[x]) {
            x = f[x] = f[f[x]];
        }
        return x;
    }
    
    bool same(int x, int y) {
        return find(x) == find(y);
    }
    
    bool merge(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return false;
        }
        siz[x] += siz[y];
        f[y] = x;
        return true;
    }
    
    int size(int x) {
        return siz[find(x)];
    }
};

void solve() {
    int n;
    std::cin >> n;

    DSU dsu(2 * n);

    std::vector<int> deg(n);
    for (int i = 0, u, v; i < n; ++i) {
        std::cin >> u >> v;
        --u, --v;
        ++deg[u], ++deg[v];

        dsu.merge(u, v + n);
        dsu.merge(v, u + n);
    }

    for (int i = 0; i < n; ++i) {
        if (deg[i] > 2 || dsu.same(i, i + n)) {
            std::cout << "NO\n";
            return;
        }
    }
    std::cout << "YES\n";
 }

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int t;
    std::cin >> t;

    while (t--) {
        solve();
    }
}

#include <bits/stdc++.h>

using i64 = long long;

void solve() {
    int n;
    std::cin >> n;
    
    std::vector<std::vector<int>> adj(n);
    for (int i = 0; i < n; i++) {
        int u, v;
        std::cin >> u >> v;
        u--;
        v--;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    
    for (int i = 0; i < n; i++) {
        if (adj[i].size() > 2) {
            std::cout << "NO\n";
            return;
        }
    }
    
    std::vector<bool> vis(n);
    for (int i = 0; i < n; i++) {
        if (!vis[i] && adj[i].size() == 2) {
            int j = i, k = adj[i][0];
            int len = 0;
            while (!vis[j] && adj[j].size() == 2) {
                vis[j] = true;
                len++;
                k ^= adj[j][0] ^ adj[j][1];
                std::swap(j, k);
            }
            
            if (j == i && len % 2 == 1) {
                std::cout << "NO\n";
                return;
            }
        }
    }
    
    std::cout << "YES\n";
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    int t;
    std::cin >> t;
    
    while (t--) {
        solve();
    }
    
    return 0;
}

F

题意：容器题，B中元素每次可以 乘除2，问B能否等于A

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思路：从二进制视角看，1是只减不增的。AB相等保证高位的1相等
消去末尾的0，AB比配，匹配不上的B消去末尾一个1，接着向下匹配。

#include <bits/stdc++.h>

using i64 = long long;

void solve() {
    int n;
    std::cin >> n;

    std::map<int, int> cnt;

    std::vector<int> a(n), b(n);

    for (int i = 0; i < n; ++i) {
        std::cin >> a[i];
        int x = a[i] >> __builtin_ctz(a[i]);
        --cnt[x];
    }
    for (int i = 0; i < n; ++i) {
        std::cin >> b[i];
        int x = b[i] >> __builtin_ctz(b[i]);
        ++cnt[x];
    }

    while (not cnt.empty()) {
        auto [k, v] = *cnt.rbegin();
        cnt.erase(cnt.rbegin()->first);

        if (v == 0) {
            continue;
        }

        if (v < 0) {
            std::cout << "NO\n";
            return;
        }
        
        int x = k >> (__builtin_ctz(k >> 1) + 1);
        cnt[x] += v;
    }

    std::cout << "YES\n";
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int t;
    std::cin >> t;

    while (t--) {
        solve();
    }
}

G

题意：

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思路：