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  • Codeforces Round #277 (Div. 2) B.OR in Matrix 模拟

    B. 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 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 mrows consisting of n integers representing matrix A that can produce given matrix B. If there are several solutions print any one.

    Sample test(s)
    input
    2 2
    1 0
    0 0
    output
    NO
     
    题意:给你一个矩阵,推原矩阵
    题解:模拟就好了
    //1085422276
    #include<bits/stdc++.h>
    using namespace std ;
    typedef long long ll;
    #define mem(a) memset(a,0,sizeof(a))
    #define pb push_back
    #define meminf(a) memset(a,127,sizeof(a));
    
    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;
    }
    //****************************************
    #define maxn 1000+5
    #define mod 1000000007
    
    int c[maxn][maxn],b[maxn][maxn],a[maxn][maxn];
    int main(){
    
    
       int n=read();
        int m=read();
        memset(b,-1,sizeof(b));
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf("%d",&a[i][j]);
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(a[i][j]==0){
                    for(int k=1;k<=m;k++){
                        b[i][k]=0;
                    }
                    for(int k=1;k<=n;k++){
                        b[k][j]=0;
                    }
                }
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(b[i][j]==-1)b[i][j]=1;
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                    int aa=0;
                    for(int k=1;k<=m;k++){
                        aa|=b[i][k];
                    }
                    for(int k=1;k<=n;k++){
                        aa|=b[k][j];
                    }
                    c[i][j]=aa;
            }
        }bool flag=0;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(c[i][j]!=a[i][j])flag=1;
            }
        }
        if(!flag){
            cout<<"YES"<<endl;
            for(int i=1;i<=n;i++){
                for(int j=1;j<=m;j++){
                    cout<<b[i][j]<<" ";
                }
                cout<<endl;
            }
        }
        else cout<<"NO"<<endl;
      return 0;
    }
    代码
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  • 原文地址:https://www.cnblogs.com/zxhl/p/4948733.html
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