双端队列o(1)模拟区间翻转

链接

#include <bits/stdc++.h>
#define IO ios::sync_with_stdio(0)
using namespace std;

const int N = 1e5 + 5;
bitset<N> v;

int main() {
    IO;
    int n; cin >> n;
    vector<int> a(n + 1), p(n + 1);
    deque<int> q;

    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        p[a[i]] = i;
    }

    int st = 1, k = 0;
    while (st <= n && a[st] == st) st++;

    if (st == n + 1) cout << "yes\n1" << endl;
    else {
    	//op记录从队头插入还是从队尾插入
        int op = 1, f = 1, k = abs(st - p[st]) + 1;
        for (int i = st; i <= n;) {
            if (i == a[i]) {
                i++;
                continue;
            }

			//如果一个数在他第一次翻转之后仍留在原地，则该情况不合法，体现为v[i] == 1
            if (v[i] || abs(i - p[i]) + 1 != k) {
                f = 0;
                break;
            }

            for (int j = i + q.size(); j < i + k; ++j) {
                if (op) q.push_front(a[j]);
                else q.push_back(a[j]);
                v[a[j]] = 1;
            }

            //for (auto j = q.begin(); j != q.end(); ++j) cout << (*j) << " ";cout << endl;

            if (op) {
                int cnt = 0;
                for (auto j = q.begin(); j != q.end(); ++j) {
                    if ((*j) == i) {
                        i++;
                        cnt++;
                    } else {
                        break;
                    }
                }
                while (cnt--) q.pop_front();
            } else {
                int cnt = 0;
                //这是从队尾插入的，所以要倒着遍历
                for (auto j = q.rbegin(); j != q.rend(); ++j) {
                    if ((*j) == i) {
                        i++;
                        cnt++;
                    } else {
                        break;
                    }
                }
                while (cnt--) q.pop_back();
            }
            op = !op;
        }
        if (q.size()) f = 0;
        A:
            if (f) cout << "yes\n" << k << endl;
            else cout << "no\n";
    }
    return 0;
}

标签：cnt,牛牛,交换,++,st,int,else,排序,op
来源： https://blog.csdn.net/weixin_45563175/article/details/115584382