C-11 problem h: treasure chest 2

Problem H: Open the treasure chest 2
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 1443 Solved: 860
Description
Pioneer is a businessman , One day I found a treasure chest , The treasure box needs the right password to open . At the same time, he found a number on the chest , And a password list . There are... On the password list n A code , Only one password is correct .

The pioneer's Island has m Locations , Each location has two mysterious numbers . He gets the information he has at each location through trading , I also know that the number on this treasure chest is the label of a place x. The vanguard needs to arrive first x Location No ,x The first number on the location number is the label of the final location he is going to , The second number on the final location is the serial number of the password on the password table .

Because the treasure in the treasure chest is too valuable . Every once in a while , The numbers at some locations will change .

The pioneer wants to know the password to open this treasure chest , Smart, can you tell him directly ？

Input
Two numbers on the first line n,m.(1<=n,m<=20)

Next n A digital ai Indicates the serial number on the password table 1 The serial number n What are your passwords .(1 <=ai<=100)

Next m Two numbers on each line u,v.(1 <= u<=m,1<= v <=n)

And then I'll give you a T, Express T operations .(1<=T<= 20)

Next T That's ok , The first number in each line op, It means the first one op Kind of operation .

The first 1 Kind of operation ： Then there's a number x, Represents the number on the treasure chest .(1<=x<=m).

The first 2 Kind of operation ： Then there are three numbers x,u,v, Express x Change the number of location to u,v.

Output
Perform the first operation each time 1 after , Output a number to indicate the last password .（ Each number accounts for 1 That's ok ）.

Sample Input
5 4
1 2 3 4 5
2 4
3 3
1 2
2 5
4
1 1
2 1 4 2
1 1
1 2
Sample Output
3
5
2

#include <stdio.h>

int main()
{
    
	int n,m;
	scanf("%d%d",&n,&m);
	int a[25][4];
	int b[25];
	for ( int i = 1;i <= n;i++)
	{
    
		scanf("%d",&b[i]);
	}
	for (int i = 1;i <= m;i++)
	{
    
		for (int j = 1;j <= 2;j++)
		{
    
			scanf("%d",&a[i][j]);
		}
	}
	int T,flag,t;
	scanf("%d",&T);
	for (int i = 0;i < T;i++)
	{
    
		scanf("%d",&flag);
		//printf("flag = %d\n",flag);
		if (flag == 1)
		{
    
			scanf("%d",&t);
			int p = a[t][1];
			int q = a[p][2];
			printf("%d\n",b[q]);
		}
		else
		{
    
			scanf("%d",&t);
			scanf("%d%d",&a[t][1],&a[t][2]);
			//printf("nnn\n");
		}
	}
	return 0;
}

This problem is actually a paper tiger , Look hard , It just looks hard . It's hard to understand that long paragraph of reading , Understand the meaning of the topic , There's no problem at all . Also, the process will be cumbersome .