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

B. OR in Matrix

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/486/problem/B

Description

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:

where 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:

.

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

Sample Input

2 2
1 0
0 0

Sample Output

NO

HINT

题意

给你b矩阵，bij = ai1 | ai2 | ai3 ...... | aim | a1j | a2j ..... | anj

然后让你求a矩阵

题解：

找找规律就知道，如果bij是0，那么a矩阵中，第i行和第j行都是0

如果bij是1，那么a矩阵中，第i行或者第j行存在一个1就好了

代码

#include<stdio.h>
#include<iostream>
using namespace std;
int a[110][110];
int vis1[110];
int vis2[110];
int main()
{
    int n,m;scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            scanf("%d",&a[i][j]);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(a[i][j]==0)
            {
                vis1[i]=1;
                vis2[j]=1;
            }
        }
    }
    int sum1=0;
    for(int i=1;i<=n;i++)
        sum1+=vis1[i];
    int sum2=0;
    for(int i=1;i<=m;i++)
        sum2+=vis2[i];
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(a[i][j]==1)
            {
                if(sum1==n||sum2==m)
                {
                    printf("NO
");
                    return 0;
                }
                if(vis1[i]&&vis2[j])
                {
                    printf("NO
");
                    return 0;
                }
            }
        }
    }
    printf("YES
");
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(vis1[i]||vis2[j])
            {
                printf("0 ");
            }
            else
                printf("1 ");
        }
        printf("
");
    }
}