【codeforces486B】OR in Matrix（思维）

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本文介绍了一个关于矩阵操作的问题——B.ORinMatrix。任务要求根据给定的矩阵B推测原始矩阵A，并验证B是否正确计算得出。文章详细阐述了解决方案的思路与实现过程。

Let's define logical OR as an operation on two logical values (i. e. values that belong to the set {0, 1}) that is equal to 1 if either or both of the logical values is set to 1, otherwise it is 0. We can define logical OR of three or more logical values in the same manner:

$\text{[math]}$ where $\text{[math]}$ is equal to 1 if some ai = 1, otherwise it is equal to 0.

Nam has a matrix A consisting of m rows and n columns. The rows are numbered from 1 to m, columns are numbered from 1 to n. Element at row i (1 ≤ i ≤ m) and column j (1 ≤ j ≤ n) is denoted as Aij. All elements of A are either 0 or 1. From matrix A, Nam creates another matrix B of the same size using formula:

$\text{[math]}$ .

(Bij is OR of all elements in row i and column j of matrix A)

Nam gives you matrix B and challenges you to guess matrix A. Although Nam is smart, he could probably make a mistake while calculating matrix B, since size of A can be large.

Input

The first line contains two integer m and n (1 ≤ m, n ≤ 100), number of rows and number of columns of matrices respectively.

The next m lines each contain n integers separated by spaces describing rows of matrix B (each element of B is either 0 or 1).

Output

In the first line, print "NO" if Nam has made a mistake when calculating B, otherwise print "YES". If the first line is "YES", then also print mrows consisting of n integers representing matrix A that can produce given matrix B. If there are several solutions print any one.

Examples

input
2 2
1 0
0 0

output
NO

input
2 3
1 1 1
1 1 1

output
YES
1 1 1
1 1 1

input
2 3
0 1 0
1 1 1

output
YES
0 0 0
0 1 0

本题就是先求出A数列，然后求出对应的正确的B，再与所给的实际的B一一对比，如果全都相同则YES，否则错误。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N = 105;
int row[N],col[N];
bool visr[N],visc[N],vis[N][N];
int a[N][N],d[N][N];
int main() {
	int n,m;
	while(scanf("%d%d",&m,&n)!=EOF) {
		for(int i=0; i<m; i++) {
			for(int j=0; j<n; j++) {
				scanf("%d",&a[i][j]);
			}
		}
		memset(row,0,sizeof(row));
		memset(col,0,sizeof(col));
		for(int i=0; i<m; i++) {
			for(int j=0; j<n; j++) {
				if(a[i][j]==0) {
					row[i]=1;
					col[j]=1;
				}
			}
		}
		int flag=0;
		memset(visr,0,sizeof(visr));
		memset(visc,0,sizeof(visc));
		for(int i=0; i<m; i++) {
			for(int j=0; j<n; j++) {
				if(!row[i]&&!col[j]) {
					vis[i][j]=1;
					visr[i]=1;
					visc[j]=1;
				}
			}
		}
		memset(d,0,sizeof(d));
		for(int i=0; i<m; i++) {
			for(int j=0; j<n; j++) {
				if(visr[i]||visc[j]) {
					d[i][j]=1;
				}
				if(a[i][j]!=d[i][j]) {
					flag=1;
				}
			}
		}
		if(flag) {
			printf("NO\n");
			continue;
		} else {
			printf("YES\n");
			for(int i=0; i<m; i++) {
				for(int j=0; j<n; j++) {
					if(j==n-1)
						printf("%d\n",vis[i][j]);
					else
						printf("%d ",vis[i][j]);
				}
			}
		}
	}
	return 0;
}