Codeforces Round #344 (Div. 2) B 题题解 （暴力）

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 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.

Examples

input

3 3 3
1 1 3
2 2 1
1 2 2

output

3 1 3
2 2 2
0 1 0

input

5 3 5
1 1 1
1 3 1
1 5 1
2 1 1
2 3 1

output

1 1 1
1 0 1
1 1 1
1 0 1
1 1 1

题意:就是在一个n行m列的矩阵中，你有一把刷子，可以对一行或者一列进行刷上颜色。且后面刷的颜色可以覆盖前面刷的颜色。大意如此，具体题意与相关参数不在此赘述。

题解：记录下行与列被刷的情况，某一点的颜色则可以确定为该点在行，与列上被刷更晚的那一个。直接输出其颜色即可。暴力即可解决；

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#include <vector>
using namespace std;

const int maxx = 5000 + 10;
int N,M;
struct node{
int color;
int k;
}rc[2][maxx];
int k;///op数

int new_max(node x,node y){
if(x.k > y.k){
return x.color;
}else{
return y.color;
}
}

int main( ){
cin>>N>>M>>k;
int op,op_count,color;
for(int i = 0;i < k; ++i){
rc[1][op_count].color = 0;
rc[1][op_count].k = 0;
rc[0][op_count].color = 0;
rc[0][op_count].k = 0;
}
for(int i = 1;i <= k; ++i){
cin>>op>>op_count>>color;
rc[op-1][op_count].color = color;
rc[op-1][op_count].k = i;
}
for(int i = 1;i <= N; ++i){
for(int  j = 1;j <= M; ++j){
printf("%d",new_max(rc[0][i],rc[1][j]));
if(j!=M) printf(" ");
}
printf("\n");
}
return 0;
}

如有BUG，欢迎指出~

联系方式：hh_0828@outlook.com 

不胜感激～