[Codeforces 486B] Or in Matrix

[题目链接]

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

[算法]

        显然 ， 当Bi,j为0时 ， 矩阵A的第i行和第j列所有数均为0

        将其它元素填充为1 ， 判断是否合法即可

        时间复杂度 ： O(N^3)

[代码]

         

#include<bits/stdc++.h>
using namespace std;
#define MAXN 110

int m , n;
int a[MAXN][MAXN],b[MAXN][MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
int main()
{
        
        read(m); read(n);
        memset(a,255,sizeof(a));
        for (int i = 1; i <= m; i++)
        {    
                for (int j = 1; j <= n; j++)
                {
                        read(b[i][j]);
                        if (b[i][j] == 0)
                        {
                                for (int k = 1; k <= n; k++) a[i][k] = 0;
                                for (int k = 1; k <= m; k++) a[k][j] = 0;    
                        }                
                }
        }
        for (int i = 1; i <= m; i++)
        {
                for (int j = 1; j <= n; j++)
                {
                        if (a[i][j] == -1)
                                a[i][j] = 1;
                }
        }
        for (int i = 1; i <= m; i++)
        {
                for (int j = 1; j <= n; j++)
                {
                        int res = 0;
                        for (int k = 1; k <= n; k++) res |= a[i][k];
                        for (int k = 1; k <= m; k++) res |= a[k][j];
                        if (res != b[i][j])
                        {
                                printf("NO
");
                                return 0;        
                        }                
                }
        }
        printf("YES
");
        for (int i = 1; i <= m; i++)
        {
                for (int j = 1; j < n; j++)
                {
                        printf("%d ",a[i][j]);
                }
                printf("%d
",a[i][n]);
        }
        
        return 0;
    
}