SDUTOJ------3306

英文题看不懂

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

b矩阵是a的或，已知b求啊；

思路：反推a，找矛盾；

题目描述

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.

输入

 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).

输出

 In the first line, print "NO" if Nam has made a mistake when calculating B, otherwise print "YES".

示例输入

2 21 00 02 31 1 11 1 1

示例输出

NOYES

提示

 

来源

 scf0920

示例程序

#include <iostream>#include<cstring>#include<cstdio>#include<cmath>#include<queue>#include<algorithm>#include<stack>#include<set>#include<vector>#define LL long longusing namespace std;int a[110][110],b[110][110];int main(){     int m,n;     while(~scanf("%d%d",&n,&m))     {         for(int i=1;i<=n;i++)         {             for(int j=1;j<=m;j++)                a[i][j]=1;         }         for(int i=1;i<=n;i++)         {             for(int j=1;j<=m;j++)             {                 scanf("%d",&b[i][j]);                 if(!b[i][j])                 {                     for(int k=1;k<=n;k++)                     {                         a[k][j]=0;                     }                     for(int k=1;k<=m;k++)                     {                         a[i][k]=0;                     }                 }             }         }         bool flag=1;         for(int i=1;i<=n;i++)         {             for(int j=1;j<=m;j++)             {                 if(b[i][j])                 {                     flag=0;                     for(int k=1;k<=n;k++)                     {                         if(a[k][j])                         {                             flag=1;                             break;                         }                     }                     for(int k=1;k<=m;k++)                     {                         if(a[i][k])                         {                             flag=1;                             break;                         }                     }                 }                 if(!flag)                 break;             }             if(!flag)                break;         }         if(flag)            printf("YES\n");         else            printf("NO\n");     }    return 0;}