Codeforces--486B---OR in Matrix思维题

OR in Matrix

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 to1, otherwise it is 0. We can define logicalOR of three or more logical values in the same manner:

where is equal to1 if some a_i = 1, otherwise it is equal to0.

Nam has a matrix A consisting of m rows and n columns. The rows are numbered from1 to m, columns are numbered from1 to n. Element at rowi (1 ≤ i ≤ m) and columnj (1 ≤ j ≤ n) is denoted asA_ij. All elements ofA are either 0 or 1. From matrix A, Nam creates another matrix B of the same size using formula:

.

(B_ij isOR of all elements in row i and column j of matrix A)

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

Input

The first line contains two integer m andn (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 either0 or 1).

Output

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

Examples

Input

2 21 00 0

Output

NO

Input

2 31 1 11 1 1

Output

YES1 1 11 1 1

Input

2 30 1 01 1 1

Output

YES0 0 00 1 0

矩阵A通过OR运算变成矩阵B，现给出矩阵B，问能不能求出矩阵A，若有多组，输出任意一组矩阵A

OR运算 ：  A[i][j] 表示矩阵A第i行第j列的所有数中包含1，则B[i][j]就是1，否则就是0

通过B矩阵反求A矩阵

#include<cstdio>#include<cstring>#include<map>#include<cmath>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define PI acos(-1.0)#define inf 0x3f3f3f3f#define mt(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long LL;const int maxn = 1e2+20;int pos[maxn][maxn], put[maxn][maxn];int main(){    int m, n, ans, flag;    scanf("%d%d", &m, &n);    mt(pos, 0); mt(put, 0);//memset只能初始化0，-1,0x3f3f3f3f这三个值，所以只能绕个圈子标记    for(int i = 0; i < m; i++){        for(int j = 0; j < n; j++){            scanf("%d", &pos[i][j]);            if(pos[i][j] == 0){//B矩阵一个值为0，则A矩阵这个值行与列都不可能出现1                for(int k = 0; k < m; k++){                    put[k][j] = 1;//用1标记0，即矩阵A不可能出现1的位置，最后1,0颠倒输出即可                }                for(int k = 0; k < n; k++){                    put[i][k] = 1;                }            }        }    }    ans = 1;    for(int i = 0; i < m; i++){        for(int j = 0; j < n; j++){//所有不可能出现1的位置已标记            if(pos[i][j] == 1){//判断所有B矩阵是1的位置，判断该行与列是否可以出现1                flag = 0;                for(int k = 0; k < m; k++){                    if(put[k][j] == 0){                        flag = 1;                        break;                    }                }                if(flag) continue;                for(int k = 0; k < n; k++){                    if(put[i][k] == 0){                        flag = 1;                        break;                    }                }                if(flag) continue;                ans = 0;                break;            }        }    }    if(ans){        printf("YES\n");        for(int i = 0; i < m; i++){            for(int j = 0; j < n; j++){                printf("%d%c", !put[i][j], j == n-1 ? '\n' : ' ');//1,0反着输出            }        }    }    else{        printf("NO\n");    }    return 0;}