Codeforces 631 B Print Check【思维】

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 from1 to n, while columns are numbered from left to right with integers from1 to m. Initially, all cells are painted in color0.

Your program has to support two operations:

1. Paint all cells in row r_i in colora_i;
2. Paint all cells in column c_i in colora_i.

If during some operation i there is a cell that have already been painted, the color of this cell also changes toa_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 rowr_i is painted in colora_i;
• 2 c_i a_i (1 ≤ c_i ≤ m,1 ≤ a_i ≤ 10⁹), means that columnc_i is painted in colora_i.

Output

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

Examples

Input

3 3 31 1 32 2 11 2 2

Output

3 1 3 2 2 2 0 1 0 

Input

5 3 51 1 11 3 11 5 12 1 12 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.

题目大意：

有一个n*m的矩阵，有k个操作，每一次操作有三个元素，第一个元素表示操作行还是列，第二个元素表示对应哪一行/哪一列，最后一个元素表示要更改的数值。

思路：

1、首先定义两个结构体，row【i】，col【i】；其中设定两个元素，val、time；一个表示这一行要修改的元素值，time表示这一行修改的时间，然后在输入的时候，不断的更新其内容。

2、那么枚举每一个格子，对应输出修改的行/列时间较晚的那一个，如果对应当前格子，行赋值较晚，那么这个格子就赋值给行修改的值，否则这个格子就赋值给列的修改的值，具体参考代码。

Ac代码:

#include<stdio.h>#include<string.h>using namespace std;struct node{    int time,val;}row[5005],col[5005];int n,m,k;int ans[5005][5005];int main(){    while(~scanf("%d%d%d",&n,&m,&k))    {        memset(row,0,sizeof(row));        memset(col,0,sizeof(col));        for(int i=0;i<k;i++)        {            int op,x,val;            scanf("%d%d%d",&op,&x,&val);            if(op==1)            {                row[x].time=i+1;                row[x].val=val;            }            else            {                col[x].time=i+1;                col[x].val=val;            }        }        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                if(row[i].time>col[j].time)                {                    ans[i][j]=row[i].val;                }                else ans[i][j]=col[j].val;            }        }        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                printf("%d ",ans[i][j]);            }               printf("\n");        }    }}