【csp模拟赛2】 序列操作

   线性推，开数组太麻烦，可以用指针

代码：

#include <iostream>
#include <cstdio>
#include <queue>
using namespace std;
const int N = 100010;
inline int read()
{
	int x = 0 , f = 1;	char ch = getchar();
	while(ch < '0' || ch > '9')	{if(ch == '-')	f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
struct node
{
	int val,vis;
	node *nxt , *fa;
	node () {val = vis = 0;nxt = fa = NULL;}
}*root;
struct num
{
	int val;
	node *p;
	friend bool operator < (const num & a,const num & b) { return a.val < b.val;}
};
int n , cnt;
priority_queue<num>q;
int main()
{
	#ifdef yilnr
	#else
	freopen("ysy.in","r",stdin);
	freopen("ysy.out","w",stdout);
	#endif
	n = read();
	node *p = root =new node();
	for(int i = 1;i <= n;i ++)
	{
		p -> val = read();
		q.push( (num) {p -> val,p});
		if(i < n)
		{
			p -> nxt = new node();
			p -> nxt -> fa = p;
			p = p -> nxt;
		}
	}
	while(cnt != (n >> 1))
	{
		node *p = q.top().p;
		q.pop();
		if(p -> nxt == NULL) continue;
		if(p -> vis || p -> nxt -> vis) continue;
		printf("%d %d ",p -> val,p -> nxt -> val);
		cnt ++;
		if(p -> fa && p -> nxt && p -> nxt -> nxt)
		{
			p -> nxt -> nxt -> fa = p -> fa;
			p -> fa -> nxt = p -> nxt -> nxt;
		} 
		p -> vis = 1;
		p -> nxt -> vis = 1;
	}
	fclose(stdin);
	fclose(stdout);
	return 0;
}
/*
12
4 3 6 10 12 11 5 8 2 1 7 9
*/