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

巴特西

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

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