Codeforces 486B OR in Matrix【水题】

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

题意：
一个新的矩阵的元素Bij为矩阵A的第i行和A的第j列的元素相互按位与运算的结果。现在给你一个矩阵B，求矩阵A。如果存在输出YES，就把矩阵打印出来，否则输出NO。

思路：

当矩阵元素是0的时候最为特殊，现将储存答案的矩阵置为1，遇到交叉点是0的情况就把矩阵交叉点所在的行列全部置为0。处理结束后再查一遍1的位置是否合法。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
//typedef long long LL;
//typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int MAXN = 1e5;
int c[110][110], b[110][110];
int main() {
    int n, m;
    while (scanf("%d%d", &n, &m) != EOF) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                scanf("%d", &b[i][j]);
                c[i][j] = 1;
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (!b[i][j]) {
                    for (int k = 0; k < m; k++) {
                        c[i][k] = 0;
                    }
                    for (int k = 0; k < n; k++) {
                        c[k][j] = 0;
                    }
                }
            }
        }
        bool flag;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                flag = false;
                if (b[i][j]) {
                    flag = true;
                    for (int k = 0; k < m; k++) {
                        if (c[i][k]) flag = false;
                    }
                    for (int k = 0; k < n; k++) {
                        if (c[k][j]) flag = false;
                    }
                }
                if (flag) break;
            }
            if (flag) break;
        }
        if (flag) printf("NO
");
        else {
            printf("YES
");
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m - 1; j++) {
                    printf("%d ", c[i][j]);
                }
                printf("%d
", c[i][m - 1]);
            }
        }
    }
    return 0;
}