POJ 3318 Matrix Multiplication(隨機化演算法)

畫船聽雨發表於2014-08-04
原文網址 : https://blog.csdn.net/fulongxu/article/details/38374323
隨機演算法

給你三個矩陣A，B，C。讓你判斷A*B是否等於C。

隨機一組資料，然後判斷乘以A，B之後是否與乘C之後相等。

很扯淡的啊，感覺這種演算法不嚴謹啊、、、

Matrix Multiplication
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 16255   Accepted: 3515

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.
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <time.h>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#define max( x, y )  ( ((x) > (y)) ? (x) : (y) )
#define min( x, y )  ( ((x) < (y)) ? (x) : (y) )
#define Mod 1000000007
#define LL long long

using namespace std;

const int maxn = 510;
int a[maxn][maxn];
int b[maxn][maxn];
int c[maxn][maxn];
int f[maxn];
int aa[maxn];
int bb[maxn];
int cc[maxn];

int main()
{
    int n;
    cin >>n;
    memset(aa, 0, sizeof(aa));
    memset(bb, 0, sizeof(bb));
    memset(cc, 0, sizeof(cc));
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++) scanf("%d",&a[i][j]);
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++) scanf("%d",&b[i][j]);
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++) scanf("%d",&c[i][j]);

    srand((unsigned int)time(0));
    for(int i = 0; i < n; i++)
        f[i] = rand()%100;
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        {
            bb[i] += b[i][j]*f[j];
            cc[i] += c[i][j]*f[j];
        }
    }
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++) aa[i] += a[i][j]*bb[j];
    int flag = 0;
    for(int i = 0; i < n; i++)
    {
        if(aa[i] != cc[i])
        {
            flag = 1;
            break;
        }
    }
    if(flag)
        cout<<"NO"<<endl;
    else
        cout<<"YES"<<endl;
}


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