Matrix multiplication（bitset）

最新推荐文章于 2021-06-16 16:51:06 发布

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#二维数组

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本文探讨如何利用bitset数据结构来优化矩阵乘法的计算，通过位操作提高矩阵乘法的效率，适用于大规模矩阵运算的场景。

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Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1928 Accepted Submission(s): 872

Problem Description

Given two matrices A and B of size n×n, find the product of them.

bobo hates big integers. So you are only asked to find the result modulo 3.

Input

The input consists of several tests. For each tests:

The first line contains n (1≤n≤800). Each of the following n lines contain n integers -- the description of the matrix A. The j-th integer in the i-th line equals A _ij. The next n lines describe the matrix B in similar format (0≤A _ij,B _ij≤10 ⁹).

Output

For each tests:

Print n lines. Each of them contain n integers -- the matrix A×B in similar format.

Sample Input

Sample Output

Author

Xiaoxu Guo (ftiasch)

Source

2014 Multi-University Training Contest 5

第一次接触bitset，纠结了好久，也光是调试就调试了二十多分钟，听说还可以用n3解决，不过要改变循环的顺序；
下面这句是抄大神的；

这样写会超时：
            for (int i=1; i<=n; i++)
                for (int j=1; j<=n; j++)
                    for (int k=1; k<=n; k++)
                        c[i][j]+=a[i][k]*b[k][j];

这样写就能过:

        for (int k=1; k<=n; k++)
            for (int i=1; i<=n; i++)
                for (int j=1; j<=n; j++)
                    c[i][j]+=a[i][k]*b[k][j];

我们知道内存中二维数组是以行为单位连续存储的，逐列访问将会每次跳1000*4(bytes)。根据cpu cache的替换策略，将会有大量的cache失效。
时间居然会相差很多。

以下是我的bitset代码：

AC 代码

/****************************************************** /
 /*                                                   /*
  *   ***********                                      *
  *           *      Auther:     ZSGG                  *
  *         *                                          *
  *       *          Name:        bitset               *
  *     *                                              *
  *   ***********    Algorithm:                        *
  *                                                    *
  */
/*******************************************************/


#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <bitset>
#include <queue>
#include <vector>
#include <cstdlib>
#include <algorithm>

#define ls u << 1
#define rs u << 1 | 1
#define lson l, mid, u << 1
#define rson mid + 1, r, u << 1 | 1

using namespace std;
const int M = 810;
int b[M][M],n;

int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    bitset<810>t[M][3]; //Every elements has eight hundrys and ten elements;
    bitset<810>s[M][3];

    while(~scanf("%d",&n))
    {

        for(int i = 1; i <= n; i++) //清空bitset容器；
        {
            for(int j = 1; j < 3; j++) 
            {
                t[i][j].reset(); 
                s[i][j].reset();
            }
        }

        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                int x;
                scanf("%d",&x);
                x %= 3;
                if(x) t[i][x][j] = 1;  //在t[i][x]这个元素的第j个位置设置为1；
            }
        }
        
        // 用bitset可以固定位置，这可以提高效率，使内层循环从n缩小为2 * 2； 详细看下面的代码；
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                scanf("%d",&b[i][j]);
                b[i][j] %= 3;
            }
        }

        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                if(b[j][i]) s[i][b[j][i]][j] = 1; //按列进bitset
            }
        }
       
        /*核心代码*/
        /*************************************************************/
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                   int res = 0;
                   for(int p = 1; p <= 2; p++) 
                   {
                       for(int k = 1; k <= 2; k++)
                       {
                           res += (t[i][p] & s[j][k]).count() * p * k; //  求出行和列的并集个数；如 0010 1010 交集个数为1；
                       }
                   }

                   printf(j == n ? "%d\n" : "%d ", res % 3);
            }
        }
        
        /************************************************************/

    }
    return 0;
}

AC 代码：

/****************************************************** /
 /*                                                   /*
  *   ***********                                      *
  *           *      Auther:     ZSGG                  *
  *         *                                          *
  *       *          Name:        bitset               *
  *     *                                              *
  *   ***********    Algorithm:                        *
  *                                                    *
  */
/*******************************************************/


#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <bitset>
#include <queue>
#include <vector>
#include <cstdlib>
#include <algorithm>

#define ls u << 1
#define rs u << 1 | 1
#define lson l, mid, u << 1
#define rson mid + 1, r, u << 1 | 1

using namespace std;
const int M = 810;
int a[M][M],b[M][M],c[M][M];

int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int n;
    while(~scanf("%d",&n))
    {
        memset(c,0,sizeof(c));
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                 { scanf("%d",&a[i][j]); a[i][j] %= 3; }
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                { scanf("%d",&b[i][j]); b[i][j] %= 3; }
        for(int k = 1; k <= n; k++)
        {
            for(int i = 1; i <= n; i++)
            {
                for(int j = 1; j <= n; j++)
                {
                    c[i][j] += b[k][j] * a[i][k];
                }
            }
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j  = 1; j <= n; j++)
            {
                if(j == n) printf("%d\n",c[i][j] % 3);
                else printf("%d ",c[i][j] % 3);
            }
        }
    }
    return 0;
}