01204212/Zooma 1

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This task is motivated by Zuma, a video game by PopCap Games.

In this version of the game, there is a sequence of n colored balls that moves toward an exit. You can shoot another m colored balls into the sequence. Balls do not disappear in this version.

Find out the final sequence of the balls.

The balls in the original sequence are numbered from 1 to n. The balls that you shoot are numbered from n+1 to n+m.

Game play example

Consider the case where n=5 and m=4. The original sequence has balls with these colors:

        1     2     3     4     5
        G     B     G     Y     Y

You have 4 balls numbered as this:

        6     7     8     9
        R     G     B     G

If you shoot your first ball to the location after ball 3, the sequence becomes

        1     2     3     *6*     4     5
        G     B     G     *R*     Y     Y

If you shoot your second ball to the location after ball 1, the sequence becomes

        1     *7*     2     3     6     4     5
        G     *G*     B     G     R     Y     Y

If you shoot your third ball to the location after ball 6, the sequence becomes

        1     7     2     3     6     *8*     4     5
        G     G     B     G     R     *B*     Y     Y

If you shoot your forth ball to the location after ball 5, the sequence becomes

        1     7     2     3     6     8     4     5     *9*
        G     G     B     G     R     B     Y     Y     *G*

and this is the final sequence.

Note: In this version, the colors of the balls are not relevant because balls do not disappear. In next versions of the task (see Zooma 2 and Zooma 3), there are rules about the disappearance of consecutive balls with the same color.

Input

• First line: n and m
• Next n lines: for 1 <= i <= n, line 1 + i specifies one integer c[i] the color of ball i.
• Next m lines: for 1 <= j <= m, line 1 + n + j specifies two integers d[j] and p[j]. d[j] is the color of ball n+j (this is your j-th ball), and p[j] is the number of the ball right after which you shoot this ball into the sequence. Note that p[j] < n + j.

• • Limits in grader:** In the grader n <= 100,000 and m <= 100,000. The time limit is 3 seconds.

Output

Your program should print n+m integers which are the ball numbers in the final sequence.

Example

This is the same example with colors: G = 1, B = 2, Y = 3, R = 4.

Input

5 4
1
2
1
3
3
4 3
1 1
2 6
1 5

Output

1
7
2
3
6
8
4
5
9

Test data

Download at: http://theory.cpe.ku.ac.th/~jittat/courses/01204212/tasks/zooma1/

Expected output for n10-3.in:

1
2
14
18
20
3
11
13
4
5
17
6
7
8
15
12
19
9
10
16

Expected output for n50-3.in (first 10 lines):

1
61
2
97
52
3
81
4
71
67
5

Code

Use the following code to get your started. The code basically reads the input. Note that the arrays are object fields, and they are 0-based (i.e., first element from the input is placed in array at index 0). So be careful about the indices (i.e., ball 1's color is kept at c[0]). Anyway, you can modify method readInput so that it processes the input the way you like.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	private int n,m;
	private int[] c;
	private int[] d;
	private int[] p;
	
	public static void main(String[] args) throws IOException {
		Main main = new Main();
		
		main.process();
	}

	private void process() throws IOException {
	    readInput();
	    // ... do something here 
	}

	private void readInput() throws IOException {
	    BufferedReader reader = new BufferedReader(
	               new InputStreamReader(System.in) );

	    String[] items = reader.readLine().split(" "); 
		n = Integer.parseInt(items[0]);
		m = Integer.parseInt(items[1]);

		c = new int[n];
		d = new int[m];
		p = new int[m];
		
		for(int i=0; i<n; i++) {
			c[i] = Integer.parseInt(reader.readLine());
		}

		for(int j=0; j<m; j++) {
			items = reader.readLine().split(" ");
			d[j] = Integer.parseInt(items[0]);
			p[j] = Integer.parseInt(items[1]);
		}
	}
}

Next challenge

Check out Zooma 2.