题目描述
blablabla
样例
#include<iostream>
const int N = 1e5+10;
int head, e[N], en[N], idx;
void init()
{
head = -1, idx = 0;
}
void sert_in_head(int x)
{
e[idx] = x, en[idx] = head, head = idx++;
}
void dlt_k_next(int k)
{
en[k] = en[en[k]];
}
void ist_in_k(int k, int x)
{
e[idx] = x, en[idx] = en[k], en[k] = idx++;
}
int main()
{
int m = 0;
std::cin>> m;
init();
while(m--)
{
int k = 0, x = 0;
char op;
std::cin >> op;
if(op == 'H')
{
std::cin >> x;
sert_in_head(x);
}
else if(op == 'D')
{
std::cin >> k;
if (k == 0)
{
head = en[head];
}
else
dlt_k_next(k-1);
}
else
{
std::cin >> k >> x;
ist_in_k(k-1, x);
}
}
for(int i = head; i != -1; i = en[i]) printf("%d ", e[i]);
printf("\n");
return 0;
}
blablabla
算法1
(暴力枚举) $O(n^2)$
blablabla
时间复杂度
参考文献
C++ 代码
blablabla
算法2
(暴力枚举) $O(n^2)$
blablabla
时间复杂度
参考文献
C++ 代码
blablabla