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最新推荐文章于 2022-03-30 14:50:48 发布

爱上键盘的小哥哥

最新推荐文章于 2022-03-30 14:50:48 发布

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本文介绍了一个二维矩阵涂色的问题及其实现方案。该方案通过记录每一行和每一列的最后涂色状态，来高效地输出整个矩阵经过一系列涂色操作后的最终状态。

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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:

Paint all cells in row r_i in color a_i;
Paint all cells in column c_i in color a_i.

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

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 r_i a_i (1 ≤ r_i ≤ n, 1 ≤ a_i ≤ 10⁹), means that row r_i is painted in color a_i;
2 c_i a_i (1 ≤ c_i ≤ m, 1 ≤ a_i ≤ 10⁹), means that column c_i is painted in color a_i.

Output

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

Example

Input

Output

3 1 3 
2 2 2 
0 1 0

Input

Output

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.

$\text{[math]}$

题意：

在n*m的方格上涂色，一共有k次操作，每次操作输入3个数字op，a，b

op==1 在第a行上涂b色

op==2 在第a列上涂b色

颜色能够覆盖，即一个方格被多次涂色后，只能看到最后最后一次涂的颜色

打印k次操作后格子的颜色状况

代码：

#include<stdio.h>
#include<string.h>
struct node{
    int a;//记录颜色
    int t;//记录被涂色的时间
}row[5005],col[5005];
int n,m,k;
int main(){
    memset(row,0,sizeof(row));
    memset(col,0,sizeof(col));
    scanf("%d%d%d",&n,&m,&k);
    int i;
    for(i=1;i<=k;i++){//注意要从1开始
        int op,p,q;
        scanf("%d%d%d",&op,&p,&q);
        if(op==1){
            row[p].a=q;
            row[p].t=i;
        }
        else{
            col[p].a=q;
            col[p].t=i;
        }
    }
    int j;
    for(i=1;i<=n;i++){  //打印结果
        int first=1;
        for(j=1;j<=m;j++){  //打印最后被涂的颜色，没被涂过输出0
            if(first) first=0;
            else printf(" ");
            if(row[i].t>col[j].t){
                printf("%d",row[i].a);
            }
            else{
                printf("%d",col[j].a);
            }
        }
        printf("\n");
    }
return 0;
}