2021 HDU多校 Boring data structure problem 链表模拟

Posted 2021-09-06 kaka0010

原题链接：https://acm.hdu.edu.cn/showproblem.php?pid=7072

目录

• 题意
• 分析
• Code

题意

有四个操作

1. 在左端插入一个数
2. 在右端插入一个数
3. 删除值为x的元素（保证唯一）
4. 查询第$(\frac{m+1}{2})$个元素的值

分析

赛中一直被删除操作困扰很久，看了题解之后恍然大悟。

思路基本就是维护两个链表，我们控制两个链表的size，使得第$\frac{m+1}{2}$个元素正好位于第二个链表的左端，这样第四个操作可以O(1)查询。然后插入时维护大小也基本是O(1)。

接着就是删除操作，我们可以先不删除，给这个数打上标记，并给当前所在的链表size-1，在移动链表或查询时，如果当前元素已经被删除了，我们直接略过这个元素，接着找下一个，直到满足条件为止，这样时间复杂度也基本是O(1)，而删除操作最多$1.5e6$，时间复杂度也不会超过$O(1e7+1.5e6)$

记得手写链表，不然会RE。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int ul;
typedef pair<int, int> PII;
const ll inf = 2e18;
const int N = 1e7 + 10;
const int M = 1e6 + 10;
const ll mod = 1e9 + 7;
const double eps = 1e-8;

#define lowbit(i) (i & -i)
#define Debug(x) cout << (x) << endl
#define fi first
#define se second
#define mem memset
#define endl '\\n'

bool vis[N];
int LSize, RSize;
int bel[N];
struct List {
    int a[N<<1];
    int l = 1e7, r = 1e7-1;
    int back() {return a[r];}
    int front() {return a[l];}
    void push_front(int x) {a[--l] = x;}
    void push_back(int x) {a[++r] = x;}
    void pop_front() {++l;}
    void pop_back() {--r;}
}L, R;
void rebuild1() {
    while (LSize > RSize) {
        int now = L.back();
        if (!vis[now]) {
            R.push_front(now), RSize++;
            bel[now] = 2;
            LSize--;
        }
        L.pop_back();
    }
}
void rebuild2() {
    while (LSize < RSize - 1) {
        int now = R.front();
        if (!vis[now]) {
            L.push_back(now), RSize--;
            bel[now] = 1;
            LSize++;
        }
        R.pop_front();
    }
}
int ask() {
    while (1) {
        int now = R.front();
        if (!vis[now]) {
            return now;
        } else {
            R.pop_front();
        }
    }
}
inline void solve() {
    int n; cin >> n;
    int cnt = 0;
    while (n--) {
        char op; cin >> op;
        if (op == 'L') {
            L.push_front(++cnt);
            bel[cnt] = 1;
            LSize++;
            rebuild1();
        } else if (op == 'R') {
            R.push_back(++cnt);
            bel[cnt] = 2;
            RSize++;
            rebuild2();
        } else if (op == 'G') {
            int x; cin >> x;
            if (bel[x] == 1) {
                LSize--;
                rebuild2();
            }
            else {
                RSize--;
                rebuild1();
            }
            vis[x] = 1;
        } else {
            printf("%d\\n", ask());
        }
        //cout << LSize << ' ' << RSize << endl;
    }
}
signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
    signed test_index_for_debug = 1;
    char acm_local_for_debug = 0;
    do {
        if (acm_local_for_debug == '$') exit(0);
        if (test_index_for_debug > 20)
            throw runtime_error("Check the stdin!!!");
        auto start_clock_for_debug = clock();
        solve();
        auto end_clock_for_debug = clock();
        cout << "Test " << test_index_for_debug << " successful" << endl;
        cerr << "Test " << test_index_for_debug++ << " Run Time: "
             << double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
        cout << "--------------------------------------------------" << endl;
    } while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
#else
    solve();
#endif
    return 0;
}