[暴力] Educational Codeforces Round 71 (Rated for Div. 2) B. Square Filling (1207B)

题目:http://codeforces.com/contest/1207/problem/B

 

B. Square Filling

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two matrices A

and B. Each matrix contains exactly n rows and m columns. Each element of A is either 0 or 1; each element of B is initially 0

.

You may perform some operations with matrix B

. During each operation, you choose any submatrix of B having size 2×2, and replace every element in the chosen submatrix with 1. In other words, you choose two integers x and y such that 1≤x<n and 1≤y<m, and then set Bx,y, Bx,y+1, Bx+1,y and Bx+1,y+1 to 1

.

Your goal is to make matrix B

equal to matrix A. Two matrices A and B are equal if and only if every element of matrix A is equal to the corresponding element of matrix B

.

Is it possible to make these matrices equal? If it is, you have to come up with a sequence of operations that makes B

equal to A

. Note that you don't have to minimize the number of operations.

Input

The first line contains two integers n

and m (2≤n,m≤50

).

Then n

lines follow, each containing m integers. The j-th integer in the i-th line is Ai,j. Each integer is either 0 or 1

.

Output

If it is impossible to make B

equal to A, print one integer −1

.

Otherwise, print any sequence of operations that transforms B

into A in the following format: the first line should contain one integer k — the number of operations, and then k lines should follow, each line containing two integers x and y for the corresponding operation (set Bx,y, Bx,y+1, Bx+1,y and Bx+1,y+1 to 1). The condition 0≤k≤2500

should hold.

Examples

Input

Copy

3 3
1 1 1
1 1 1
0 1 1

Output

Copy

3
1 1
1 2
2 2

Input

Copy

3 3
1 0 1
1 0 1
0 0 0

Output

Copy

-1

Input

Copy

3 2
0 0
0 0
0 0

Output

Copy

0

Note

The sequence of operations in the first example:

000000000→110110000→110110110→110111111

题意:

给两个n*m的01矩阵A和B,给出了A中的元素，B中元素全为０，现在有一个操作可以在B中选一个坐标，它和右边下面右下的元素都变为１，问能否通过一些操作（不要求最少）使得B等于A，若能则输出步数和坐标，否则输出-１

思路:

暴力并判断边界(当时忘记判断边界结果被Hack了 ~TAT~ )

 #include<bits/stdc++.h>
 using namespace std;
 const int amn=;
 int a[amn][amn],idx[amn][amn],ansx[],ansy[];
 int main(){
     int n,m,tp=,valid=;
     cin>>n>>m;
     for(int i=;i<=n;i++){
         for(int j=;j<=m;j++){
             cin>>a[i][j];
             idx[i][j]=;
         }
     }
     for(int i=;i<=n;i++){
         for(int j=;j<=m;j++){
             if(a[i][j]){
                 if(!idx[i][j]&&(i+>n||j+>m)){ ///这里判断边界条件被hack了...当时想着i<n&&j<m,判非法时忘记判边界了
                     valid=0;
                     break;
                 }
                 if(a[i][j+]&&a[i+][j]&&a[i+][j+]){
                     idx[i][j+]=idx[i+][j]=idx[i+][j+]=;
                     ansx[++tp]=i;ansy[tp]=j;
                 }
                 else if(!idx[i][j]&&(!idx[i][j+]||!idx[i+][j]||!idx[i+][j+])){
                     valid=;
                     break;
                 }
             }
         }
         if(valid==)break;
     }
     if(valid){
         cout<<tp<<endl;
         for(int i=;i<=tp;i++)cout<<ansx[i]<<' '<<ansy[i]<<endl;
     }
     else cout<<-<<endl;
 }
 /**
 给两个n*m的01矩阵A和B,给出了A中的元素，B中元素全为０，现在有一个操作可以在B中选一个坐标，它和右边下面右下的元素都变为１，问能否通过一些操作（不要求最少）使得B等于A，若能则输出步数和坐标，否则输出-１
 暴力并判断边界
 **/