Codeforces Round #277 (Div. 2)B——OR in Matrix

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#codeforces

探讨了一个名为B.ORinMatrix的编程问题，该问题要求根据特定逻辑运算规则逆向推导原始矩阵A，从给定的矩阵B出发，验证输入是否合理并给出可能的解决方案。

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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 a_i = 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 A_ij. All elements of A are either 0 or 1. From matrix A, Nam creates another matrix B of the same size using formula:

$\text{[math]}$ .

(B_ij 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 m rows consisting of n integers representing matrix A that can produce given matrix B. If there are several solutions print any one.

Sample test(s)

我也不知道题解是怎么做的，反正我是胡搞搞出来的

#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int m, n;
int matb[105][105];
int mat[105][105];
int c[105][105];

int main()
{

    while(~scanf("%d%d", &m, &n))
    {
        int ans1 = 0;
        memset (matb, 0, sizeof(matb));
        memset (c, 0,sizeof(c));
        for (int i = 1; i <= m; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                scanf("%d", &mat[i][j]);
                matb[i][j] = mat[i][j];
                if (mat[i][j] == 1)
                {
                    ans1++;
                }
                else
                {
                    c[i][j] = 1;
                }
            }
        }
        for (int i = 1; i <= m; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (c[i][j] == 1)
                {
                    for (int x = 1; x <= m; x++)
                    {
                        if (matb[x][j])
                        {
                            matb[x][j] = 0;
                        }
                    }
                    for (int y = 1; y <= n; y++)
                    {
                        if (matb[i][y])
                        {
                            matb[i][y] = 0;
                        }
                    }
                }
            }
        }
        bool flag =  false;
        for (int i = 1; i <= m; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                int c = 0;
                for (int x = 1; x <= m; x++)
                {
                    c |= matb[x][j];
                }
                for (int y = 1; y <= n; y++)
                {
                    c |= matb[i][y];
                }
                if (c != mat[i][j])
                {
                    flag = true;
                    break;
                }
            }
            if (flag)
            {
                break;
            }
        }
        if (flag)
        {
            printf("NO\n");
            continue;
        }
        printf("YES\n");
        for (int i = 1; i <= m; i++)
        {
            for (int j = 1; j < n; j++)
            {
                printf("%d ",matb[i][j]);
            }
            printf("%d\n", matb[i][n]);
        }
    }
    return 0;
}