源代码:ACM/OpenjudgeNow/Codeforces at master · abmcar/ACM (github.com)

更好的阅读体验: 折跃坐标

A. Playoff

题目大意:

在这里插入图片描述

思路:

比较简单的数学思路

我们发现每轮会刷掉奇数位的选手

不难发现和证明 最终剩下的一定是$2^n-1 $

代码:

void solution()
{
    cin >> n;
    cout << (1 << n) - 1 << endl;
}

B. Prove Him Wrong

题目大意:

在这里插入图片描述

思路:

根据上图,我们可以发现当数组为a~i~,3*a~i~时,操作不会减少

因此,我们可以构造出$1,3,9...$这样的数组,判断$a_i\geq 1e9$即可

代码:

void solution()
{
    cin >> n;
    vector<ll> V;
    ll nowNum = 1;
    while (V.size() < n)
    {
        V.push_back(nowNum);
        nowNum *= 3;
        if (nowNum > 1e9)
            break;
    }
    if (V.size() == n)
    {
        cout << "YES" << endl;
        for (auto it : V)
            cout << it << " ";
        cout << endl;
    }
    else
    {
        cout << "NO" << endl;
    }
}

C. Fault-tolerant Network

题目大意:

在这里插入图片描述

思路:

一个恶心人的思路题罢了

有这么几种连法

  • 连$(a_1,b_1),(a_n,b_n)$

  • 连$(a_1,b_1),(a_1,b_n),(a_n,b_x)$,其中x时任意位置

  • 连$(a_1,b_),(a_n,b_),(a_,b_1),(a_,b_n)$其中x,y均为任意位置

建议自己画个图好好比划一下

代码:

void solution()
{
    ll ans = 1e12;
    cin >> n;
    vector<ll> ga(n), gb(n);
    for (int i = 0; i < n; i++)
        cin >> ga[i];
    for (int i = 0; i < n; i++)
        cin >> gb[i];
    ans = min({ans, abs(ga[0] - gb[0]) + abs(ga[n - 1] - gb[n - 1]), abs(ga[0] - gb[n - 1]) + abs(ga[n - 1] - gb[0])});
    {
        ll tmpAns = abs(ga[0] - gb[0]);
        ll t1 = 1e12;
        ll t2 = 1e12;
        for (int i = 1; i < n; i++)
            t1 = min(t1, abs(ga[n - 1] - gb[i]));
        for (int i = 1; i < n; i++)
            t2 = min(t2, abs(gb[n - 1] - ga[i]));
        tmpAns = tmpAns + t1 + t2;
        ans = min(ans, tmpAns);
    }
    {
        ll tmpAns = abs(ga[n - 1] - gb[n - 1]);
        ll t1 = 1e12;
        ll t2 = 1e12;
        for (int i = 0; i < n - 1; i++)
            t1 = min(t1, abs(ga[0] - gb[i]));
        for (int i = 0; i < n - 1; i++)
            t2 = min(t2, abs(gb[0] - ga[i]));
        tmpAns = tmpAns + t1 + t2;
        ans = min(ans, tmpAns);
    }
    {
        ll tmpAns = abs(ga[0] - gb[n - 1]);
        ll t1 = 1e12;
        ll t2 = 1e12;
        for (int i = 0; i < n - 1; i++)
            t1 = min(t1, abs(ga[n - 1] - gb[i]));
        for (int i = 1; i < n; i++)
            t2 = min(t2, abs(gb[0] - ga[i]));
        tmpAns = tmpAns + t1 + t2;
        ans = min(ans, tmpAns);
    }
    {
        ll tmpAns = abs(ga[n - 1] - gb[0]);
        ll t1 = 1e12;
        ll t2 = 1e12;
        for (int i = 1; i < n; i++)
            t1 = min(t1, abs(ga[0] - gb[i]));
        for (int i = 0; i < n - 1; i++)
            t2 = min(t2, abs(gb[n - 1] - ga[i]));
        tmpAns = tmpAns + t1 + t2;
        ans = min(ans, tmpAns);
    }
    {
        ll tmpAns = abs(ga[n - 1] - gb[0]) + abs(ga[n - 1] - gb[n - 1]);
        ll t1 = 1e12;
        for (int i = 1; i < n; i++)
            t1 = min(t1, abs(ga[0] - gb[i]));
        tmpAns = tmpAns + t1;
        ans = min(ans, tmpAns);
    }
    {
        ll tmpAns = abs(gb[0] - ga[0]) + abs(gb[0] - ga[n - 1]);
        ll t1 = 1e12;
        for (int i = 1; i < n; i++)
            t1 = min(t1, abs(gb[n - 1] - ga[i]));
        tmpAns = tmpAns + t1;
        ans = min(ans, tmpAns);
    }
    {
        ll tmpAns = abs(gb[n - 1] - ga[0]) + abs(gb[n - 1] - ga[n - 1]);
        ll t1 = 1e12;
        for (int i = 1; i < n; i++)
            t1 = min(t1, abs(gb[0] - ga[i]));
        tmpAns = tmpAns + t1;
        ans = min(ans, tmpAns);
    }
    {
        ll tmpAns = abs(ga[0] - gb[0]) + abs(ga[0] - gb[n - 1]);
        ll t1 = 1e12;
        for (int i = 1; i < n; i++)
            t1 = min(t1, abs(ga[n - 1] - gb[i]));
        tmpAns = tmpAns + t1;
        ans = min(ans, tmpAns);
    }
    {
        ll tmpAns = 0;
        ll t1 = 1e12;
        ll t2 = 1e12;
        for (int i = 1; i < n - 1; i++)
            t1 = min(t1, abs(ga[0] - gb[i]));
        for (int i = 1; i < n - 1; i++)
            t2 = min(t2, abs(ga[n - 1] - gb[i]));
        tmpAns = tmpAns + t1 + t2;
        t1 = t2 = 1e12;
        for (int i = 1; i < n - 1; i++)
            t1 = min(t1, abs(gb[0] - ga[i]));
        for (int i = 1; i < n - 1; i++)
            t2 = min(t2, abs(gb[n - 1] - ga[i]));
        tmpAns = tmpAns + t1 + t2;
        ans = min(ans, tmpAns);
    }
    cout << ans << endl;
}

D. Nearest Excluded Points

题目大意:

在这里插入图片描述

思路:

很容易可以想到一种从已有点进行bfs的做法,但会超时,

如果这2e5个点组成了一个正方形或者圆形,时间复杂度接近$O(n^2)$

正难则反,在脑子里模拟一遍flood fill,可以发现里已有点最近的点总是在已有点附近,那么我们可以从重点出发,向内bfs寻找起点,如此一来,时间复杂度降为$O(n\log n)$

代码:

代码使用了部分c++2a新增功能

int n;
    cin >> n;
    map<pair<int, int>, int> vis, mp;
    set<pair<int, int>> s;
    vector<pair<int, int>> ans(n + 1);
    for (int i = 1; i <= n; i++)
    {
        int t1, t2;
        cin >> t1 >> t2;
        mp[{t1, t2}] = i;
        for (auto [dx, dy] : dir4)
            s.insert({t1 + dx, t2 + dy});
    }
    queue<array<int, 5>> q;
    for (auto [x, y] : s)
        if (!mp.contains({x, y}))
            q.push({x, y, x, y, 0});
    while (!q.empty())
    {
        auto [orix, oriy, x, y, dis] = q.front();
        q.pop();
        for (auto [dx, dy] : dir4)
        {
            if (vis[{x + dx, y + dy}] == 1)
                continue;
            if (!mp.contains({x + dx, y + dy}))
                continue;
            {
                ans[mp[{x + dx, y + dy}]] = {orix, oriy};
                q.push({orix, oriy, x + dx, y + dy, dis + 1});
                vis[{x + dx, y + dy}] = 1;
            }
        }
    }

    for (int i = 1; i <= n; i++)
    {
        auto [x, y] = ans[i];
        cout << x << " " << y << '\n';
    }