HDU 4920 Matrix multiplication(矩陣相乘)

畫船聽雨發表於2014-08-05
矩陣

各種TEL，233啊。沒想到是處理掉0的情況就可以過啊。一直以為會有極端資料。沒想到竟然是這樣的啊、、在網上看到了一個AC的神奇的程式碼，經典的矩陣乘法，只不過把最內層的列舉，移到外面就過了啊、、、有點不理解啊，複雜度不是一樣的嗎、、

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 640    Accepted Submission(s): 250


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 Aij. The next n lines describe the matrix B in similar format (0≤Aij,Bij≤109).
 

Output
For each tests:

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

Sample Input
1
0
1
2
0 1
2 3
4 5
6 7





 



Sample Output




0
0 1
2 1





 



Author



Xiaoxu Guo (ftiasch)



 



Source



2014 Multi-University Training Contest
5


#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-12
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

using namespace std;


const int maxn = 810;

int a[maxn][maxn];
int b[maxn][maxn];
int c[maxn][maxn];
int aa[maxn][maxn];
int bb[maxn][maxn];

int main()
{
    int n;
    while(cin >>n)
    {
        memset(c, 0, sizeof(c));
        memset(aa, 0, sizeof(aa));
        memset(bb, 0, sizeof(bb));
        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 i = 1; i <= n; i++)
        {
            int x = -1;
            for(int j = n; j >= 0; j--)
            {
                aa[i][j] = x;
                if(a[i][j]) x = j;
            }
        }

        for(int i = 1; i <= n; i++)
        {
            int x = -1;
            for(int j = n; j >= 0; j--)
            {
                bb[i][j] = x;
                if(b[i][j]) x = j;
            }
        }
        for (int i = 1; i <= n; i++)
        {
            for(int j = aa[i][0]; j != -1; j = aa[i][j])
            {
                for(int k = bb[j][0]; k != -1; k = bb[j][k])
                    c[i][k] += a[i][j]*b[j][k];
            }
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n-1; j++)
                printf("%d ",c[i][j]%3);
            printf("%d\n",c[i][n]%3);
        }
    }
    return 0;
}



這是看到有人交的AC的程式碼：





#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N = 805;
int a[N][N], b[N][N], ans[N][N];
int main()
{
    int n, i, j, k;
    while(~scanf("%d",&n))
    {
        for(i = 1; i <= n; i++)
            for(j = 1; j <= n; j++)
            {
                scanf("%d",&a[i][j]);
                a[i][j] %= 3;
            }
        for(i = 1; i <= n; i++)
            for(j = 1; j <= n; j++)
            {
                scanf("%d",&b[i][j]);
                b[i][j] %= 3;
            }
        memset(ans, 0, sizeof(ans));
        for(k = 1; k <= n; k++) //經典演算法中這層迴圈在最內層，放最內層會超時，但是放在最外層或者中間都不會超時，不知道為什麼
            for(i = 1; i <= n; i++)
                for(j = 1; j <= n; j++)
                {
                    ans[i][j] += a[i][k] * b[k][j];
                    //ans[i][j] %= 3;   //如果在這裡對3取餘，就超時了
                }
        for(i = 1; i <= n; i++)
        {
            for(j = 1; j < n; j++)
                printf("%d ", ans[i][j] % 3);
            printf("%d\n", ans[i][n] % 3);
        }
    }
    return 0;
}








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