数学（矩阵乘法，随机化算法）：POJ 3318 Matrix Multiplication

Matrix Multiplication

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17783		Accepted: 3845

Description

You are given three n × n matrices A, B 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 A, B and C 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(n³) algorithm will get TLE.

　　用个一维随机矩阵去乘再判断是否相等，类似与哈希的思想。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;
const int maxn=250010;
struct Array{
    int a[maxn],L;
    int *operator[](int x){
        return &a[(x-1)*L];
    }
};
struct Matrix{
    int R,C;
    Array mat;
    Matrix(){
        memset(mat.a,0,sizeof(mat.a));
        R=C=0;
    }
    void Init(int r,int c){
        R=r;mat.L=C=c;
    }
    int *operator[](int x){
        return mat[x];
    }
    friend Matrix operator*(Matrix a,Matrix b){
        Matrix c;c.Init(a.R,b.C);
        for(int i=1;i<=a.R;i++)
            for(int j=1;j<=b.C;j++)
                for(int k=1;k<=a.C;k++)
                    c[i][j]+=a[i][k]*b[k][j];
        return c;            
    }
    friend bool operator ==(Matrix a,Matrix b){
        if(a.R!=b.R||a.C!=b.C)return false;
        for(int i=1;i<=a.R;i++)
            for(int j=1;j<=b.C;j++)
                if(a[i][j]!=b[i][j])
                    return false;
        return true;            
    }
}A,B,C,D;
int main(){
    int n,x;
    scanf("%d",&n);srand(n);
    A.Init(n,n);
    B.Init(n,n);
    C.Init(n,n);
    D.Init(n,1);
    for(int i=1;i<=n;i++)
        D[i][1]=rand();
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            scanf("%d",&x);
            A[i][j]=x;
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            scanf("%d",&x);
            B[i][j]=x;
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            scanf("%d",&x);
            C[i][j]=x;
        }
    }
    A=A*(B*D);
    C=C*D;
    if(A==C)
        printf("YES
");
    else
        printf("NO
");
    return 0;    
}