AcWing 4654. 消除游戏

        首先看范围  说明这题应该在  范围的时间复杂度，那么想到一共最多n个字符需要删除，且保证删除时间在  。于是想到用双链表来实现删除操作。还有很多细节见代码。

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 10;

int n;
char str[N];
int l[N], r[N];
vector <int> w, q;
bool visited[N];

void insert(int x)
{
    if (!visited[x])
    {
        visited[x] = true;
        w.push_back(x);
    }
}

void Initialization()
{
    w.clear();
    
    for (int k : q)
    {
        int a = l[k], b = k, c = r[k];
        if (str[a] == str[b] && str[b] != str[c] && c != n + 1) 
        {
            insert(b);
            insert(c);
        }
        if (str[a] != str[b] && str[b] == str[c] && a != 0)
        {
            insert(a);
            insert(b);
        }
    }
}

int main()
{
    scanf("%s", str + 1);
    n = strlen(str + 1);
    str[0] = str[n + 1] = '@';
    
    for (int i = 1; i <= n; ++ i)
    {
        l[i] = i - 1, r[i] = i + 1;
        q.push_back(i);
    }
    l[n + 1] = n, r[0] = 1;
    
    while (true)
    {
        Initialization();
        if (w.empty()) break;
        
        q.clear();
        for (int k : w)
        {
            int a = l[k], b = k, c = r[k];
            if (!visited[a] && a != 0 && (q.empty() ||q.back() != a)) q.push_back(a);
            if (!visited[c] && c != n + 1) q.push_back(c);
            l[c] = a, r[a] = c;
        }
    }
    
    if (r[0] == n + 1) puts("EMPTY");
    else 
    {
        for (int i = r[0]; i != n + 1; i = r[i])
            cout << str[i];
    }
    
    return 0;
}