码迷,mamicode.com
首页 > 其他好文 > 详细

OR in Matrix

时间:2017-03-30 11:19:05      阅读:225      评论:0      收藏:0      [点我收藏+]

标签:equal   ted   output   inf   desc   inpu   puts   ons   spec   

OR in Matrix
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
Submit Status

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:

技术分享 where 技术分享 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:

技术分享.

(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

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:定义一个异或,a1^a2...^an如果有一个ai=1那么值为1,否则为零,给出你一个矩阵B,是由矩阵A得来的,Bij等于A的i行元素
    异或j列元素。给出你矩阵B问你是否有这样的矩阵A

初步思路:将矩阵初始化为1,然后先按照矩阵B中有零的元素,将对应A矩阵中的元素设置成零,然后在反过来验证B矩阵
*/
#include <bits/stdc++.h>
using namespace std;
int n,m;
int a[110][110];
int b[110][110];
int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            scanf("%d",&b[i][j]);
            a[i][j]=1;
        }
    }
    //按照B矩阵进行置0
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(b[i][j]==0){
                for(int k=1;k<=m;k++)
                    a[i][k]=0;
                for(int k=1;k<=n;k++)
                    a[k][j]=0;
            }
        }
    }
    //验证然后按照1的位置验证B矩阵
    bool f=false;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(b[i][j]==1){
                bool flag=false;
                for(int k=1;k<=n;k++){
                    if(a[i][k]==1){
                        flag=true;
                        break;
                    }
                }
                if(flag==false)
                for(int k=1;k<=m;k++){
                    if(a[k][j]==1){
                        flag=true;
                        break;
                    }
                }
                if(flag==false){
                    f=true;
                    break;
                }
            }
        }
    }
    if(f){
        puts("NO");
    }else{
        puts("YES");
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                printf(j==1?"%d":" %d",a[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}

 

OR in Matrix

标签:equal   ted   output   inf   desc   inpu   puts   ons   spec   

原文地址:http://www.cnblogs.com/wuwangchuxin0924/p/6644262.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!