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  • POJ 3318 Matrix Multiplication (随机化)

    You are given three n × n matrices AB and C. Does the equation A × B = C hold true?

    Input

    The first line of input contains a positive integer n (n ≤ 500) followed by the the three matrices AB and respectively. Each matrix's description is a block of n × n integers.

    It guarantees that the elements of A and B are less than 100 in absolute value and elements of C are less than 10,000,000 in absolute value.

    Output

    Output "YES" if the equation holds true, otherwise "NO".

    Sample Input

    2
    1 0
    2 3
    5 1
    0 8
    5 1
    10 26
    

    Sample Output

    YES

    Hint

    Multiple inputs will be tested. So O(n 3) algorithm will get TLE.
    题解:
      给你三个n*n的矩阵A、B、C,问你A*B是不是等于C。数据保证O(n^3)铁定超时,因此你需要想一个不用把A和B乘起来就可以验证的算法。一个基于概率的算法是随机生成一个n乘1的矩阵R,然后判断A*B*R是否等于C*R,而前者相当于A*(B*R),与后者一样都可以在O(n^2)的时间里算出来。如果算出来的结果相等,几乎可以肯定A*B和C也是相等的。
    C++:
    #include<iostream>
    #include<ctime>
    #include<cstring>
    #include<cstdio>
    #include<cstdlib>
    using namespace std;
    const int maxn=505;
    int r[maxn],A[maxn][maxn],B[maxn][maxn],C[maxn][maxn],rA[maxn],rAB[maxn],rC[maxn];
    int n;
    void input(int mat[maxn][maxn])
    {
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                scanf("%d",&mat[i][j]);
    }
    bool check()
    {
        for(int j=0;j<n;j++)
            for(int i=0;i<n;i++)
            {
                rA[j]+=r[i]*A[i][j];
                rC[j]+=r[i]*C[i][j];
            }
        for(int j=0;j<n;j++)
            for(int i=0;i<n;i++)
                rAB[j]+=rA[i]*B[i][j];
        for(int i=0;i<n;i++)
            if(rAB[i]!=rC[i])
                return false;
        return true;
    }
    int main()
    {
        cin>>n;
        srand(time(NULL));
        input(A);
        input(B);
        input(C);
        for(int i=0;i<n;i++)
            r[i]=rand()%99+1;
        if(check())
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
        return 0; 
    }
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  • 原文地址:https://www.cnblogs.com/orion7/p/7358731.html
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