稀疏矩阵的乘法

#include<iostream>
using namespace std;
struct Data
{
    int row,col;
    int num;
};
class Matrix
{
private:
    int M[3][4],N[4][3];
    struct Data data1[10],data2[10];//这是一种数据类型
    int k,p;//k用来记录有多少个有效数值。
public:
    void getdata()
    {
        for(i=0; i<3; i++)
        {
            for(j=0; j<4; j++)
            {
                cin>>M[i][j];
            }
        }
        for(i=0; i<4; i++)
        {
            for(j=0; j<3; j++)
            {
                cin>>N[i][j];
            }
        }

    }
    void getnew()
    {
        int i,j;
        k=0;
        p=0;
        for(i=0; i<3; i++)
        {
            for(j=0; j<4; j++)
            {
                if(M[i][j]!=0)
                {
                    data1[k].row=i;
                    data1[k].col=j;
                    data1[k].num=M[i][j];
                    k++;
                }
            }
        }
        for(i=0; i<4; i++)
        {
            for(j=0; j<3; j++)
            {
                if(N[i][j]!=0)
                {
                    data2[p].row=i;
                    data2[p].col=j;
                    data2[p].num=N[i][j];
                    p++;
                }
            }
        }
    }
    void result()
    {
        int re[3][3]={0};
        for(int i=0; i<k; i++)
        {
            for(int j=0; j<p; j++)
            {
                if(data1[i].col==data2[j].row)//当存在可以进行相乘的数时,还是从定义出发的
                {
                    if(re[data1[i].row][data2[j].col]==0)
                    {
                        re[data1[i].row][data2[j].col]=data1[i].num*data2[j].num;
                        //cout<<"!"<<data1[i].num*data2[j].num<<endl;
                    }

                    else re[data1[j].row][data2[i].col]=re[data1[j].row][data2[i].col]+data1[i].num*data2[j].num;
                }
            }
        }
        cout<<endl;
        for(int i=0; i<3; i++)
        {
            for(int j=0; j<3; j++)
            {
                if(re[i][j]!=0)cout<<i<<" "<<j<<" "<<re[i][j]<<endl;
            }
        }
    }
};
int main()
{
    Matrix a1;
    a1.getdata();
    a1.getnew();
    a1.result();
    return 0;
}

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转载自blog.csdn.net/qq_40554649/article/details/80098804
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