Codeforces Round #344 (Div. 2) B. Print Check

B. Print Check

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Kris works in a large company "Blake Technologies". As a best engineer of the company he was assigned a task to develop a printer that will be able to print horizontal and vertical strips. First prototype is already built and Kris wants to tests it. He wants you to implement the program that checks the result of the printing.

Printer works with a rectangular sheet of paper of size n × m. Consider the list as a table consisting of n rows and m columns. Rows are numbered from top to bottom with integers from 1 to n, while columns are numbered from left to right with integers from 1 to m. Initially, all cells are painted in color 0.

Your program has to support two operations:

1. Paint all cells in row ri in color ai;
2. Paint all cells in column ci in color ai.

If during some operation i there is a cell that have already been painted, the color of this cell also changes to ai.

Your program has to print the resulting table after k operation.

Input

The first line of the input contains three integers n, m and k (1  ≤  n,  m  ≤ 5000, n·m ≤ 100 000, 1 ≤ k ≤ 100 000) — the dimensions of the sheet and the number of operations, respectively.

Each of the next k lines contains the description of exactly one query:

• 1 ri ai (1 ≤ ri ≤ n, 1 ≤ ai ≤ 109), means that row ri is painted in color ai;
• 2 ci ai (1 ≤ ci ≤ m, 1 ≤ ai ≤ 109), means that column ci is painted in color ai.

Output

Print n lines containing m integers each — the resulting table after all operations are applied.

Examples

input

3 3 3
1 1 3
2 2 1
1 2 2

output

3 1 3 
2 2 2 
0 1 0 

input

5 3 5
1 1 1
1 3 1
1 5 1
2 1 1
2 3 1

output

1 1 1 
1 0 1 
1 1 1 
1 0 1 
1 1 1 

Note

The figure below shows all three operations for the first sample step by step. The cells that were painted on the corresponding step are marked gray.

题目要求你根据操作给矩阵染色，关键就是对于同一行（同一列）之后的操作会覆盖之前的操作，所以逆序染色标记一下，前边的就不用染了。

package codeforces344;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * Created by lenovo on 2016-03-10.
 *
 */
public class B344 {

    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;
    B344() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
        br.close();
    }
    public static void main(String[] args) throws  IOException{
        new B344();
    }

    /*
    * solve method
    * */
    /*
    * 来个逆向，就可以少对很多单元进行赋值，就只需要简单判断一下就好
    * 因为多个操作可能对应于同一行或者是同一列，所以，之后的操作肯定会覆盖之前的操作，
    * 所以就把那些无效的操作都筛掉。
    * */
    int[][] graph = new int[5000][5000];
    void solve() throws IOException {
        int n, m, k;
        n = nextInt();
        m = nextInt();
        k = nextInt();
        int[] op = new int[k + 10];
        int[] rc = new int[k + 10];
        int[] color = new int[k + 10];
        int[] fr = new int[5000 + 10];
        int[] fc = new int[5000 + 10];

        for(int i = 0; i < k; ++i) {
            op[i] = nextInt();
            rc[i] = nextInt();
            rc[i] -= 1;
            color[i] = nextInt();
        }

        for(int i = k-1; i>= 0; --i) {

            if(op[i] == 1) {
                if(fr[rc[i]] == 0){
                    fr[rc[i]] = 1;
                    for(int j = 0; j < m; ++j){
                        if(graph[rc[i]][j] == 0)
                            graph[rc[i]][j] = color[i];
                    }
                }

            } else {
                if(fc[rc[i]] == 0){
                    fc[rc[i]] = 1;

                    for(int j = 0; j < n; ++j){
                        if(graph[j][rc[i]] == 0)
                            graph[j][rc[i]] = color[i];
                    }
                }
            }
        }
        print(n, m);
    }
    
    void print(int n, int m){
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < m; ++j) {
                System.out.print(graph[i][j] + " ");
            }
            System.out.println();
        }
    }

    /*
    * 优化的流
    * */
    String nextToken(){
        while(st == null || !st.hasMoreTokens()){
            try{
                st = new StringTokenizer(br.readLine());
            } catch(IOException e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }

    String nextString(){
        try{
            return br.readLine();
        } catch (Exception e) {
            eof = true;
            return null;
        }
    }

    int nextInt() throws  IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}