「CF486B」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 from1 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 matrixA, 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 or1).
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.
|
1 2 3 |
2 2 1 0 0 0 |
|
1 |
NO |
|
1 2 3 |
2 3 1 1 1 1 1 1 |
|
1 2 3 |
YES 1 1 1 1 1 1 |
|
1 2 3 |
2 3 0 1 0 1 1 1 |
|
1 2 3 |
YES 0 0 0 0 1 0 |
题解
容易发现。。。如果矩阵的某个值为0，则原矩阵整行整列都是0
如果是1的话，那行或列上至少有1个1
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#include<iostream> #include<cstring> #include<cstdio> #include<cstdlib> #include<algorithm> #include<queue> #include<cmath> #include<map> #include<queue> #define inf 2000000000 #define ll long long using namespace std; 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; } int n,m; int a[105][105],b[105][105]; int x[105],y[105]; int main() { memset(x,1,sizeof(x)); memset(y,1,sizeof(y)); n=read();m=read(); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) a[i][j]=read(),b[i][j]=1; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { x[i]&=a[i][j]; y[j]&=a[i][j]; } for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) if(!x[i]||!y[j])b[i][j]=0; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) if(a[i][j]) { bool flag=0; for(int k=1;k<=n;k++)if(b[k][j])flag=1; for(int k=1;k<=m;k++)if(b[i][k])flag=1; if(!flag){printf("NO");return 0;} } puts("YES"); for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) printf("%d ",b[i][j]); printf("\n"); } return 0; } |
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