codeforces 631B Print Check

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.

Sample Input

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 

Sample Output

Hint

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.

题目意思很简单，就是模拟，问题是暴力会超时。

一开始我是用了很复杂的方法（当然有些正确方法的影子），主要是正确方法的一个点没想到。

什么点呢？一开始我纠结的就是一个点如果被涂了多次怎么办，一直做不出来。

正确做法就是，每次涂色都记录步数，并更新当前行/列的值，最后比较一下就可以了（注意没有涂过的情况）

下面贴代码

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;int a[5010][5010];struct myk{int num;int bianhao;};int m,n,k;myk zuizhonghang[5010],zuizhonglie[5010];void paint(int cixu,int x,int i,int j){    if(x==1)    {        zuizhonghang[i].num=j;        zuizhonghang[i].bianhao=cixu;    }    if(x==2)    {        zuizhonglie[i].num=j;        zuizhonglie[i].bianhao=cixu;    }}int main(void){    int aa,b,c;    scanf("%d%d%d",&n,&m,&k);    memset(a,0,sizeof(a));    for(int i=1;i<=5000;i++)    {        zuizhonghang[i].bianhao=0;        zuizhonglie[i].bianhao=0;    }    for(int i=1;i<=k;i++)    {        scanf("%d%d%d",&aa,&b,&c);        paint(i,aa,b,c);    }    for(int i=1;i<=n;i++)    {        for(int j=1;j<=m;j++)        {            if((zuizhonghang[i].bianhao==0)&&(zuizhonglie[j].bianhao==0))                a[i][j]=0;            else                a[i][j]=(zuizhonghang[i].bianhao>zuizhonglie[j].bianhao)?zuizhonghang[i].num:zuizhonglie[j].num;            printf("%d ",a[i][j]);        }        printf("\n");    }    return 0;}