結果

問題 No.1768 The frog in the well knows the great ocean.
ユーザー fuppy_kyopro
提出日時 2021-11-26 22:53:23
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 5,146 bytes
コンパイル時間 2,494 ms
コンパイル使用メモリ 218,292 KB
最終ジャッジ日時 2025-01-26 01:39:59
ジャッジサーバーID
（参考情報） 		judge4 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 25 WA * 2
権限があれば一括ダウンロードができます

ソースコード

diff # 

/*
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//*/

#include <bits/stdc++.h>

// #include <atcoder/all>
// #include <atcoder/maxflow>

using namespace std;
// using namespace atcoder;

#define DEBUG(x) cerr << #x << ": " << x << endl;
#define DEBUG_VEC(v)                                        \
    cerr << #v << ":";                                      \
    for (int iiiiiiii = 0; iiiiiiii < v.size(); iiiiiiii++) \
        cerr << " " << v[iiiiiiii];                         \
    cerr << endl;
#define DEBUG_MAT(v)                                \
    cerr << #v << endl;                             \
    for (int iv = 0; iv < v.size(); iv++) {         \
        for (int jv = 0; jv < v[iv].size(); jv++) { \
            cerr << v[iv][jv] << " ";               \
        }                                           \
        cerr << endl;                               \
    }
typedef long long ll;
// #define int ll

#define vi vector<int>
#define vl vector<ll>
#define vii vector<vector<int>>
#define vll vector<vector<ll>>
#define vs vector<string>
#define pii pair<int, int>
#define pis pair<int, string>
#define psi pair<string, int>
#define pll pair<ll, ll>
template <class S, class T>
pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) {
    return pair<S, T>(s.first + t.first, s.second + t.second);
}
template <class S, class T>
pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template <class S, class T>
ostream &operator<<(ostream &os, pair<S, T> p) {
    os << "(" << p.first << ", " << p.second << ")";
    return os;
}
#define X first
#define Y second
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define rrep(i, n) for (int i = (int)(n)-1; i >= 0; i--)
#define rrep1(i, n) for (int i = (int)(n); i > 0; i--)
#define REP(i, a, b) for (int i = a; i < b; i++)
#define in(x, a, b) (a <= x && x < b)
#define all(c) c.begin(), c.end()
void YES(bool t = true) {
    cout << (t ? "YES" : "NO") << endl;
}
void Yes(bool t = true) { cout << (t ? "Yes" : "No") << endl; }
void yes(bool t = true) { cout << (t ? "yes" : "no") << endl; }
void NO(bool t = true) { cout << (t ? "NO" : "YES") << endl; }
void No(bool t = true) { cout << (t ? "No" : "Yes") << endl; }
void no(bool t = true) { cout << (t ? "no" : "yes") << endl; }
template <class T>
bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
const long double pi = 3.1415926535897932384626433832795028841971L;
int popcount(ll t) { return __builtin_popcountll(t); }
// int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
// int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
vi dx = {0, 1, 0, -1}, dy = {-1, 0, 1, 0};
// vi dx2 = { 1,1,0,-1,-1,-1,0,1 }, dy2 = { 0,1,1,1,0,-1,-1,-1 };
struct Setup_io {
    Setup_io() {
        ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
        cout << fixed << setprecision(25);
    }
} setup_io;
// const ll MOD = 1000000007;
const ll MOD = 998244353;
// #define mp make_pair
//#define endl '\n'

void solve() {
    int n;
    cin >> n;
    vi a(n);
    rep(i, n) cin >> a[i];
    vi b(n);
    rep(i, n) cin >> b[i];

    rep(i, n) {
        if (a[i] > b[i]) {
            No();
            return;
        }
    }

    int l = 0;
    using P = pair<int, pii>;
    map<int, vector<pii>> mp;
    while (l < n) {
        int r = l;
        while (r < n && b[r] == b[l]) {
            r++;
        }
        mp[b[l]].emplace_back(l, r);
        l = r;
    }

    map<int, vi> idxs;
    rep(i, n) {
        idxs[a[i]].push_back(i);
    }

    set<int> used;
    for (auto itr = mp.begin(); itr != mp.end(); itr++) {
        int x = itr->first;
        for (auto [l, r] : itr->second) {
            auto itrr = used.lower_bound(r);
            int r2;
            if (itrr == used.end())
                r2 = n;
            else
                r2 = *itrr;

            itrr = used.lower_bound(l);
            int l2;
            if (itrr == used.begin())
                l2 = -1;
            else {
                itrr--;
                l2 = *itrr;
            }

            bool flag = false;
            auto itrrr = lower_bound(all(idxs[x]), l2 + 1);
            if (itrrr != idxs[x].end() && *itrrr < r2) {
                flag = true;
            }
            // for (int i = l2 + 1; i < r2; i++) {
            //     if (a[i] == x) {
            //         flag = true;
            //     }
            // }
            if (!flag) {
                No();
                return;
            }
        }

        for (auto [l, r] : itr->second) {
            REP(i, l, r)
            used.insert(i);
        }
    }
    Yes();
}

signed main() {
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}
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