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

B. Print Check

题目连接：

http://www.codeforces.com/contest/631/problem/B

Description

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 ri in color ai;

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.

Sample Input

3 3 3

1 1 3

2 2 1

1 2 2

Sample Output

3 1 3

2 2 2

0 1 0

Hint

题意

给你一个n*m一开始全是0的矩阵，然后又q次询问

每次询问给你三个字母 op,a,b

将第a行变成b

将第a列变成b

然后让你输出Q次询问后，这个矩阵长什么模样

题解：

记录一下每一行每一列最后被变成了什么，以及时间戳

然后我们在判断这个格子最后是啥的时候，只用看他所在的这一行和这一列的时间戳哪个比较晚就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 5005;
pair<int,int> r[maxn],c[maxn];

int main()
{
    int n,m,k;
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=n;i++)
        r[i]=make_pair(0,0);
    for(int i=1;i<=m;i++)
        c[i]=make_pair(0,0);
    for(int i=1;i<=k;i++)
    {
        int x,y,z;scanf("%d%d%d",&x,&y,&z);
        if(x==1)
            r[y]=make_pair(z,i);
        else
            c[y]=make_pair(z,i);
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(r[i].first==0&&c[j].first==0)
                printf("0 ");
            else if(r[i].first==0)
                printf("%d ",c[j].first);
            else if(c[j].first==0)
                printf("%d ",r[i].first);
            else if(r[i].second>c[j].second)
                printf("%d ",r[i].first);
            else
                printf("%d ",c[j].first);
        }
        printf("\n");
    }
}