Codeforces Round #344 (Div. 2) 631 B. Print Check （实现）

B. Print Check

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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

inputCopy

3 3 3

1 1 3

2 2 1

1 2 2

outputCopy

3 1 3

2 2 2

0 1 0

inputCopy

5 3 5

1 1 1

1 3 1

1 5 1

2 1 1

2 3 1

outputCopy

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.

B. Print Check

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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

inputCopy

3 3 3

1 1 3

2 2 1

1 2 2

outputCopy

3 1 3

2 2 2

0 1 0

inputCopy

5 3 5

1 1 1

1 3 1

1 5 1

2 1 1

2 3 1

outputCopy

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.

题意：

给你一个初始为0的n*m数组，和k个操作，每一个操作把一行或者一列变为指定的数字。让你输出最后的结果

思路：

用两个pair<int,int> 数组 r和l来记录当前行或列在第几次操作被更新，以及被更新的值是多少。

输出答案的时候记录行和列的时间戳哪个比较靠后更新就输出谁，没有记录就是0.

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int a[5003][5003];
int n,m,k;
pii r[maxn];
pii c[maxn];
int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    gbtb;
    cin>>n>>m>>k;
    int op,x,y;
    repd(i,1,k)
    {
        cin>>op>>x>>y;
        if(op==1)
        {
            r[x].fi=y;
            r[x].se=i;
        }else
        {
            c[x].fi=y;
            c[x].se=i;
        }
    }
    repd(i,1,n)
    {
        repd(j,1,m)
        {
            if(r[i].se>c[j].se)
            {
                cout<<r[i].fi<<" ";
            }else if(r[i].se<c[j].se)
            {
                cout<<c[j].fi<<" ";
            }else
            {
                cout<<0<<" ";
            }
        }
        cout<<endl;
    }
    return 0;

}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}