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

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 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.
題解：每個操作將一行或一列的數都改為同一個數。輸出最後的cells。類比一下就可以啦。直接暴力會TLE23。。。試過了。。。

AC代碼：

#pragma comment(linker, "/STACK:102400000,102400000")//#include<bits/stdc++.h>#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<map>#include<cmath>#include<queue>#include<set>#include<stack>#include <utility> using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair<int,int> pii;typedef vector<int> vi;#define mst(a) memset(a, 0, sizeof(a))#define M_P(x,y) make_pair(x,y)#define in freopen("in.txt","r",stdin) #define rep(i,j,k) for (int i = j; i <= k; i++)  #define per(i,j,k) for (int i = j; i >= k; i--)  #define lson  l , mid , rt << 1  #define rson  mid + 1 , r , rt << 1 | 1  const int lowbit(int x) { return x&-x; }  const double eps = 1e-8;  const int INF = 1e9+7; const ll inf =(1LL<<62) ;const int MOD = 1e9+7;  const ll mod = (1LL<<32);const int N =1e5+7; const int M=100010;const ll MAX=1e18;//const int maxn=1001; template <class T1, class T2>inline void getmax(T1 &a, T2 b) {if (b>a)a = b;}  template <class T1, class T2>inline void getmin(T1 &a, T2 b) {if (b<a)a = b;}int read(){int v = 0, f = 1;char c =getchar();while( c < 48 || 57 < c ){if(c=='-') f = -1;c = getchar();}while(48 <= c && c <= 57) v = v*10+c-48, c = getchar();return v*f;}ll r[5010],c[5010],s[N];int main(){ios::sync_with_stdio(false);ll n,i,j,k,m,x,b;cin>>n>>m>>k;for(i=1;i<=k;i++){cin>>x>>b>>s[i];if(x==1) r[b-1]=i;else c[b-1]=i;}for(i=0;i<n;i++){for(j=0;j<m;j++){     cout<<s[max(r[i],c[j])]<<" ";//當前位置只取決於最後一個變化 }cout<<endl;}return 0;}