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=;
 struct Array{
     int a[maxn],L;
     int *operator[](int x){
         return &a[(x-)*L];
     }
 };
 struct Matrix{
     int R,C;
     Array mat;
     Matrix(){
         memset(mat.a,,sizeof(mat.a));
         R=C=;
     }
     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=;i<=a.R;i++)
             for(int j=;j<=b.C;j++)
                 for(int k=;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=;i<=a.R;i++)
             for(int j=;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,);
     for(int i=;i<=n;i++)
         D[i][]=rand();
     for(int i=;i<=n;i++){
         for(int j=;j<=n;j++){
             scanf("%d",&x);
             A[i][j]=x;
         }
     }
     for(int i=;i<=n;i++){
         for(int j=;j<=n;j++){
             scanf("%d",&x);
             B[i][j]=x;
         }
     }
     for(int i=;i<=n;i++){
         for(int j=;j<=n;j++){
             scanf("%d",&x);
             C[i][j]=x;
         }
     }
     A=A*(B*D);
     C=C*D;
     if(A==C)
         printf("YES\n");
     else
         printf("NO\n");
     return ;
 }