• OR in Matrix


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

    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("
    ");
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/wuwangchuxin0924/p/6644262.html
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