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.
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).
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.
2 2
1 0
0 0
NO
2 3
1 1 1
1 1 1
YES
1 1 1
1 1 1
2 3
0 1 0
1 1 1
YES
0 0 0
0 1 0
题意是一个矩阵B的b[i][j]是所有A矩阵的a[i][k]和a[k][j]或起来的值,给一个B矩阵,问是否存在这样的A矩阵,并输出方案
因为是或……所以在B中出现的0必须在A中一横一竖都为0
所以先把B中0的情况搞完,然后判一下现在的B矩阵中1的位置对应的A的一行一列是否存在至少一个1
15分钟……有些慢了,你看卓神6分钟A掉
#include<cstdio> #include<iostream> #include<cstring> #include<cstdlib> #include<algorithm> #include<cmath> #include<queue> #include<deque> #include<set> #include<map> #include<ctime> #define LL long long #define inf 0x7ffffff #define pa pair<int,int> #define pi 3.1415926535897932384626433832795028841971 using namespace std; int mat[110][110]; int a[110][110]; int n,m; inline LL read() { LL x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } void ex() { printf("NO"); exit(0); } int main() { n=read();m=read(); for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) mat[i][j]=1; for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) a[i][j]=read(); for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) { int x=a[i][j]; if (!x) { for (int k=1;k<=n;k++) mat[k][j]=0; for (int k=1;k<=m;k++) mat[i][k]=0; } } for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) if(a[i][j]) { bool mrk=0; for (int k=1;k<=n;k++)if (mat[k][j])mrk=1; for (int k=1;k<=m;k++)if (mat[i][k])mrk=1; if (!mrk)ex(); } printf("YES\n"); for (int i=1;i<=n;i++) { for (int j=1;j<=m;j++) printf("%d ",mat[i][j]); printf("\n"); } }