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

B. OR in Matrix

 

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

Sample test(s)

input

2 2
1 0
0 0

output

NO

 

题意：给你一个矩阵，推原矩阵

题解：模拟就好了

//
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define meminf(a) memset(a,127,sizeof(a));

inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){
        if(ch=='-')f=-;ch=getchar();
    }
    while(ch>=''&&ch<=''){
        x=x*+ch-'';ch=getchar();
    }return x*f;
}
//****************************************
#define maxn 1000+5
#define mod 1000000007

int c[maxn][maxn],b[maxn][maxn],a[maxn][maxn];
int main(){

   int n=read();
    int m=read();
    memset(b,-,sizeof(b));
    for(int i=;i<=n;i++){
        for(int j=;j<=m;j++){
            scanf("%d",&a[i][j]);
        }
    }
    for(int i=;i<=n;i++){
        for(int j=;j<=m;j++){
            if(a[i][j]==){
                for(int k=;k<=m;k++){
                    b[i][k]=;
                }
                for(int k=;k<=n;k++){
                    b[k][j]=;
                }
            }
        }
    }
    for(int i=;i<=n;i++){
        for(int j=;j<=m;j++){
            if(b[i][j]==-)b[i][j]=;
        }
    }
    for(int i=;i<=n;i++){
        for(int j=;j<=m;j++){
                int aa=;
                for(int k=;k<=m;k++){
                    aa|=b[i][k];
                }
                for(int k=;k<=n;k++){
                    aa|=b[k][j];
                }
                c[i][j]=aa;
        }
    }bool flag=;
    for(int i=;i<=n;i++){
        for(int j=;j<=m;j++){
            if(c[i][j]!=a[i][j])flag=;
        }
    }
    if(!flag){
        cout<<"YES"<<endl;
        for(int i=;i<=n;i++){
            for(int j=;j<=m;j++){
                cout<<b[i][j]<<" ";
            }
            cout<<endl;
        }
    }
    else cout<<"NO"<<endl;
  return ;
}

代码