这题教训主要是观察和思考
本来上来的想法就是链表加类似插排?但肉眼可见的tle....
#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<sstream>
#include<string>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<map>
#include<queue>
#include<limits.h>
#include<climits>
#include<fstream>
#include<stack>
typedef long long ll;
using namespace std;
struct linknode
{
linknode* next;
ll val;
linknode* prev;
linknode() { val = 0; next = NULL; prev = NULL; }
};
bool is_not_del(linknode* s)
{
linknode* rb = s;
if (rb->next->val < rb->val)return false;
rb = rb->next;
while (rb->next != s)
{
if (rb->next->val < rb->val)return false;
rb = rb->next;
}
return true;
}
void xg(linknode* s)
{
linknode* p = s->next;
linknode* pp = p->next;
while (p != NULL)
{
delete p;
p = pp;
pp = pp->next;
}
return;
}
int main()
{
ios::sync_with_stdio(false);//imp
cin.tie(nullptr);//imp
linknode* head;
linknode* h = new linknode;
head = h;
int t; cin >> t;
for (int iii = 0; iii < t; iii++)
{
int n; cin >> n;
int xx;
cin >> xx;
h->val = xx;
for (int ii = 1; ii < n; ii++)
{
cin >> xx;
linknode* th = new linknode;
h->next = th;
th->prev = h;
h = h->next;
h->val = xx;
}
head->prev = h;
h->next = head;
int mintimes = 0;
while (true)
{
ll thval = head->val;
while (h->val >= thval and h!=head)h = h->prev;
if (h == head)break;
//插入节点
if (h == head->prev)
{
head = head->next;
h = head->prev;
}
else
{
h->next->prev = head;
linknode* headn = h->next;
linknode* hh = head->next;
h->next = head;
head->next->prev = head->prev;
head->prev->next = head->next;
head->prev = h;
head->next = headn;
head = hh;
h = head->prev;
}
mintimes++;
}
if (is_not_del(head))cout << mintimes << endl;
else cout << -1 << endl;
}
return 0;
}
但是通过观察和推理可以知道,当且仅当该列的最小元素后面有序的时候,这个才一定成立
因为每一次前面的插排到后面都不会改变原有的状态,如果后面不是满足要求的时候,那么插排再多次也没用
代码:
#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<sstream>
#include<string>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<map>
#include<queue>
#include<limits.h>
#include<climits>
#include<fstream>
#include<stack>
typedef long long ll;
using namespace std;
ll lst[200010];
int main()
{
int t; cin >> t;
for (int ii = 0; ii < t; ii++)
{
int n; cin >> n;
ll minx = LLONG_MAX;
ll minxid = 0;
for (int i = 0; i < n; i++)
{
cin >> lst[i];
if (lst[i] < minx)
{
minxid = i;
minx = lst[i];
}
}
bool is_end = true;
for (int i = minxid; i < n-1; i++)
{
if (lst[i + 1] < lst[i])
{
is_end = false;
break;
}
}
if (is_end)cout << minxid << endl;
else cout << -1 << endl;
}
return 0;
}
总结:做题要带
标签:Sort,head,prev,int,next,Queue,include,linknode From: https://www.cnblogs.com/zzzsacmblog/p/18088392