关于并查集的区间合并的思考

在下午的训练中,一道区间合并的题目卡了两个多小时,想想就可怕,回想就是思路有了,应该如何实现,却没有了具体执行的代码,在自己的一次次尝试中,也没有过去,自己之后搜索题解,发现主要就是右边界和跨域,但是自己在场上的时候却写不出来,可怕可怕。

Knight Tournament

 

Hooray! Berl II, the king of Berland is making a knight tournament. The king has already sent the message to all knights in the kingdom and they in turn agreed to participate in this grand event.

As for you, you're just a simple peasant. There's no surprise that you slept in this morning and were late for the tournament (it was a weekend, after all). Now you are really curious about the results of the tournament. This time the tournament in Berland went as follows:

  • There are n knights participating in the tournament. Each knight was assigned his unique number — an integer from 1 to n.
  • The tournament consisted of m fights, in the i-th fight the knights that were still in the game with numbers at least li and at most ri have fought for the right to continue taking part in the tournament.
  • After the i-th fight among all participants of the fight only one knight won — the knight number xi, he continued participating in the tournament. Other knights left the tournament.
  • The winner of the last (the m-th) fight (the knight number xm) became the winner of the tournament.

You fished out all the information about the fights from your friends. Now for each knight you want to know the name of the knight he was conquered by. We think that the knight number b was conquered by the knight number a, if there was a fight with both of these knights present and the winner was the knight number a.

Write the code that calculates for each knight, the name of the knight that beat him.

Input

The first line contains two integers nm (2 ≤ n ≤ 3·105; 1 ≤ m ≤ 3·105) — the number of knights and the number of fights. Each of the following m lines contains three integers li, ri, xi (1 ≤ li < ri ≤ nli ≤ xi ≤ ri) — the description of the i-th fight.

It is guaranteed that the input is correct and matches the problem statement. It is guaranteed that at least two knights took part in each battle.

Output

Print n integers. If the i-th knight lost, then the i-th number should equal the number of the knight that beat the knight number i. If the i-th knight is the winner, then the i-th number must equal 0.

Examples

Input

4 3
1 2 1
1 3 3
1 4 4

Output

3 1 4 0 

Input

8 4 3 5 4 3 7 6 2 8 8 1 8 1

Output

0 8 4 6 4 8 6 1 

Note

Consider the first test case. Knights 1 and 2 fought the first fight and knight 1 won. Knights 1 and 3 fought the second fight and knight 3 won. The last fight was between knights 3 and 4, knight 4 won.

思路:就是并查集的区间合并。

如果直接暴力的话,只有俩个for也会超时,瞬间沮丧,一直在想怎么优化两个for,陷入死循环。

在失败后,就开始想怎么精简,就在想怎么简化中间已经确定的点,直接pass掉。结合之前做过的一道不是很熟悉的区间合并的题目,所以在题解之后,自己有重新整理了思路

ac代码:

#include
#include
#include
#include
#include
#include
#include
#include
#define maxn 500005 
using namespace std;
int f[maxn];
int b[maxn];
int next[maxn];
int n,m;
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		int x,y,z,i,j;
		memset(b,0,sizeof(b)); 
		for(i=1;i<=n;i++)
		{
			f[i]=i;
			next[i]=i+1;//初始化为下一位 
		}	
		for(i=1;i<=m;i++)
		{
			scanf("%d%d%d",&x,&y,&z);
			for(j=x;j

自己写的那个,不知道为啥总是超时,哭唧唧。

 

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