Atcoder Beginner Contest 260 Tutorial

By xiaruize

A - A Unique Letter

签到题

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// Problem: A - A Unique Letter
// Contest: AtCoder - AtCoder Beginner Contest 260
// URL: https://atcoder.jp/contests/abc260/tasks/abc260_a
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define ll long long
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl;
#define NO cout << "NO" << endl;
#define multitest(t)                                                                                                   \
    while (t--)                                                                                                        \
        solve();
#define double long double
const int INF = 0x3f3f3f3f;
const int Mod = 1000000007;
const int N = 1e5 + 10;

inline int rd()
{
    int x = 0, w = 1;
    char ch = 0;
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            w = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        x = x * 10 + (ch - '0');
        ch = getchar();
    }
    return x * w;
}
inline void wt(int x)
{
    static int sta[35];
    int top = 0;
    do
    {
        sta[top++] = x % 10, x /= 10;
    } while (x);
    while (top)
        putchar(sta[--top] + 48);
}

// bool st;
string s;
// bool en;

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    cin >> s;
    if (s[0] != s[1] && s[0] != s[2])
        cout << s[0] << endl;
    else if (s[1] != s[0] && s[1] != s[2])
        cout << s[1] << endl;
    else if (s[2] != s[0] && s[1] != s[2])
        cout << s[2] << endl;
    else
        cout << "-1" << endl;
    return 0;
}

B - Better Students Are Needed!

排序基础题，写三个cmp即可

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// Problem: B - Better Students Are Needed!
// Contest: AtCoder - AtCoder Beginner Contest 260
// URL: https://atcoder.jp/contests/abc260/tasks/abc260_b
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int n, x, y, z;
struct node
{
    int m, e;
    int sum;
    int id;
    bool st;
} s[N];
// bool en;

bool cmp(node a, node b)
{
    if (a.m != b.m)
        return a.m > b.m;
    else
        return a.id < b.id;
}

bool cmpp(node a, node b)
{
    if (a.e != b.e)
        return a.e > b.e;
    else
        return a.id < b.id;
}

bool cmppp(node a, node b)
{
    if (a.sum != b.sum)
        return a.sum > b.sum;
    else
        return a.id < b.id;
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> n >> x >> y >> z;
    for (int i = 1; i <= n; i++)
        cin >> s[i].m;
    for (int i = 1; i <= n; i++)
    {
        cin >> s[i].e;
        s[i].sum = s[i].e + s[i].m;
        s[i].id = i;
        s[i].st = false;
    }
    vector<int> res;
    sort(s + 1, s + n + 1, cmp);
    int cnt = 0, p = 1;
    while (cnt < x && p <= n)
    {
        while (s[p].st)
            p++;
        s[p].st = 1;
        cnt++;
        res.pb(s[p].id);
        p++;
    }
    sort(s + 1, s + n + 1, cmpp);
    cnt = 0, p = 1;
    while (cnt < y && p <= n)
    {
        while (s[p].st)
            p++;
        s[p].st = 1;
        cnt++;
        res.pb(s[p].id);
        p++;
    }
    sort(s + 1, s + n + 1, cmppp);
    cnt = 0, p = 1;
    while (cnt < z && p <= n)
    {
        while (s[p].st)
            p++;
        s[p].st = 1;
        cnt++;
        res.pb(s[p].id);
        p++;
    }
    sort(ALL(res));
    for (auto v : res)
        cout << v << endl;
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

C - Changing Jewels

模拟题，显然最大值就是一直操作到无法操作的值

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// Problem: C - Changing Jewels
// Contest: AtCoder - AtCoder Beginner Contest 260
// URL: https://atcoder.jp/contests/abc260/tasks/abc260_c
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int n, x, y;
// bool en;

int dfs(bool st, int lv)
{
    // cerr << st << ' ' << lv << endl;
    if (lv == 1)
    {
        if (st)
            return 0;
        else
            return 1;
    }
    if (st)
    {
        int res = 0;
        res += dfs(st, lv - 1);
        res += x * dfs(0, lv);
        return res;
    }
    else
    {
        int res = 0;
        res += dfs(1, lv - 1);
        res += y * dfs(0, lv - 1);
        return res;
    }
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> n >> x >> y;
    cout << dfs(1, n) << endl;
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

D - Draw Your Cards

明显可以使用set完成加减和查询比x小的最大元素的操作，模拟即可

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// Problem: D - Draw Your Cards
// Contest: AtCoder - AtCoder Beginner Contest 260
// URL: https://atcoder.jp/contests/abc260/tasks/abc260_d
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int n, k, p[200005], res[200005];
int tp[200005], ys[200005];
set<int> s;
vector<int> st[200005];
// bool en;

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> n >> k;
    memset(res, -1, sizeof(res));
    for (int i = 1; i <= n; i++)
    {
        cin >> p[i];
        if (s.empty())
        {
            tp[i] = p[i];
            ys[p[i]] = i;
            st[i].pb(p[i]);
            s.insert(p[i]);
        }
        else
        {
            auto it = s.lower_bound(p[i]);
            if (it == s.end())
            {
                tp[i] = p[i];
                ys[p[i]] = i;
                st[i].pb(p[i]);
                s.insert(p[i]);
            }
            else
            {
                ys[p[i]] = ys[*it];
                tp[ys[p[i]]] = p[i];
                st[ys[p[i]]].pb(p[i]);
                s.erase(it);
                s.insert(p[i]);
            }
        }
        if (st[ys[p[i]]].size() == k)
        {
            for (auto x : st[ys[p[i]]])
                res[x] = i;
            st[ys[p[i]]].clear();
            s.erase(tp[ys[p[i]]]);
        }
    }
    for (int i = 1; i <= n; i++)
        cout << res[i] << endl;
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

E - At Least One

处理一个数组CiC_i表示如果不选择第i位及以前的数，右边界至少为多少。
可以O(n)处理,CA[i]=max(CA[i],B[i])C_{A[i]}=max(C_{A[i]},B[i])
然后可以枚举区间的起点，差分处理答案

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// Problem: E - At Least One
// Contest: AtCoder - AtCoder Beginner Contest 260
// URL: https://atcoder.jp/contests/abc260/tasks/abc260_e
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int n, m;
pii a[N];
int p[N];
int c[N];
int mi = INF;
// bool en;

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> a[i].fir >> a[i].sec;
    int j = 0;
    for (int i = 1; i <= n; i++)
    {
        c[a[i].fir] = max(c[a[i].fir], a[i].sec);
        j = max(j, a[i].fir);
        mi = min(a[i].sec, mi);
    }
    for (int i = 1; i <= mi; i++)
    {
        p[j - i + 1]++;
        p[m - i + 2]--;
        j = max(j, c[i]);
    }
    int res = 0;
    for (int i = 1; i <= m; i++)
    {
        res += p[i];
        cout << res << ' ';
    }
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

F - Find 4-cycle

似乎暴力就能过，这是枚举的时候可以记录一下。。。

wtcl
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// Problem: F - Find 4-cycle
// Contest: AtCoder - AtCoder Beginner Contest 260
// URL: https://atcoder.jp/contests/abc260/tasks/abc260_f
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 3e5 + 10;

// bool st;
int s, t, m;
vector<int> g[N];
int mp[3005][3005];
// bool en;

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> s >> t >> m;
    for (int i = 1; i <= m; i++)
    {
        int u, v;
        cin >> u >> v;
        g[u].pb(v - s);
    }
    for (int i = 1; i <= s; i++)
    {
        for (auto &j : g[i])
        {
            for (auto &l : g[i])
            {
                if (j == l)
                    continue;
                if (mp[j][l])
                {
                    cout << mp[j][l] << ' ' << j + s << ' ' << i << ' ' << l + s << endl;
                    return 0;
                }
                else
                    mp[j][l] = i;
            }
        }
    }
    cout << "-1" << endl;
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

理论上，时间复杂度有可能达到O(s⋅t2)O(s \cdot t^2)，但是实际上跑的飞快qwq

未完待更