Poj 3318 Matrix Multiplication( 矩阵压缩)

Matrix Multiplication

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 18928		Accepted: 4074

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.

Source

POJ Monthly--2007.08.05, qzc

 

/*
题意：有3个n*n的矩阵A,B,C，问AB是否等于C。
思路：题目描述很简单，就是用矩阵乘法，但是很明显矩阵乘法的时间复杂度为O(n^3)，很明显超时。那怎么改进呢？就是用压缩矩阵的方法：
设矩阵R是1*n的矩阵，根据矩阵的性质，若R*A*B=R*C,那么A*B=C。由此可以看出来，虽然多成了一个矩阵，但是时间复杂度成了O(n^2)。那么问题是这个R的行列式该怎么设定，有人用的随机算法，但是随机算法可能在关键点上出现错误，可以将R设定成一个递增的数列｛1，2，3……｝。
*/
#include<iostream>
#include<vector>
#include<stdio.h>
#include<string.h>
using namespace std;
int n,x;
int a[][],b[][],c[][];
bool work()
{
    int R[],ra[],rab[],rc[];
    for(int i=;i<=n;i++) {R[i]=i; ra[i]=; rab[i]=; rc[i]=;}

    for(int j=;j<=n;j++)
        for(int i=;i<=n;i++)
          ra[j]+=R[i]*a[i][j];

     for(int j=;j<=n;j++)
        for(int i=;i<=n;i++)
          rab[j]+=ra[i]*b[i][j];

     for(int j=;j<=n;j++)
        for(int i=;i<=n;i++)
          rc[j]+=R[i]*c[i][j];

     for(int i=;i<=n;i++)
        if (rab[i]!=rc[i]) return ;
    return ;
}
int main()
{
    scanf("%d",&n);
    for(int i=;i<=n;i++)
         for(int j=;j<=n;j++)
           scanf("%d",&a[i][j]);
    for(int i=;i<=n;i++)
        for(int j=;j<=n;j++)
            scanf("%d",&b[i][j]);
    for(int i=;i<=n;i++)
        for(int j=;j<=n;j++)
            scanf("%d",&c[i][j]);
    if (work()) printf("YES\n");
      else printf("NO\n");
    return ;
}