基础题 CodeForces - 631B 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 sizen × m. Consider the list as a table consisting ofn rows andm columns. Rows are numbered from top to bottom with integers from1 ton, while columns are numbered from left to right with integers from1 tom. 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 afterk operation.

Input

The first line of the input contains three integersn,m andk (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 containingm 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.

题目意思：原始矩阵为0，每个操作都会给某一行或某一列涂上一种数字，最后输出整个矩阵。

复杂度有1e5*5000,若暴力会超时。仔细想一下发现其实对于每个点（x， y），决定其最终状态的是最后x行y列的改变，因此只需求出对于每个点最终改变它的行或列即可得出答案。

解此题关键在于推敲每个操作前后会有什么变化，根据状态找出最优解法。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int clo[100000+20];int r[100000+20], l[100000+20];int main(){    int n, m, k;    scanf("%d%d%d", &n, &m, &k);    int step, wh;    memset(clo, 0, sizeof(clo));    for(int i = 1; i <= k; i++) {        scanf("%d%d%d", &step, &wh, &clo[i]);        if(step == 1)            r[wh-1] = i;        else            l[wh-1] = i;    }    for(int i = 0; i < n; i++) {        for(int j = 0; j < m; j++) {            if(j != m-1)                printf("%d ", clo[max(r[i], l[j])]);            else                printf("%d\n", clo[max(r[i], l[j])]);        }    }    return 0;}