行逻辑链接的顺序表实现稀疏矩阵的相乘(Java语言描述)

行逻辑链接,带行链接信息。程序有空指针BUG,至今未解决。还是C/C++适合描述算法数据结构。以后复杂的算法还是改用C/C++吧。

有BUG的代码,总有一天会换成没有BUG的。

package 行逻辑链接的顺序表实现稀疏矩阵的相乘;

public class Triple<T> {
	int row,col;
	T v;
	public Triple(){}
	public Triple(int row,int col, T v){
		this.row = row;
		this.col = col;
		this.v = v;
	}	
}


 

package 行逻辑链接的顺序表实现稀疏矩阵的相乘;

public class Mat {
	final int MAXSIZE = 10;
	int mu,nu,tu;
	int rpos[] = new int[MAXSIZE + 1];//各行第一个非零元的位置表
	Triple<Integer> data[] = new Triple[MAXSIZE + 1];//Java不支持泛型数组
	public Mat(int mu,int nu,int tu){
		this.mu = mu;
		this.nu = nu;
		this.tu = tu;
		for(int i=1; i<=MAXSIZE; i++)
			data[i] = new Triple();
	}
	//三元组矩阵的输出
	public void display(){
		int i,j,k,m,n,count = 0;
		for(i=1; i<=mu; i++){
			for(j=1; j<=nu; j++){
				for(k=1; k<=tu; k++){
					if(i==data[k].row && j==data[k].col){
						System.out.print(data[k].v + " ");
						count = -1;
						break;
					}
				}
				if(count != -1)
					System.out.print("0 ");
				count = 0;
			}
			System.out.println();
		}
	}
}


 

package 行逻辑链接的顺序表实现稀疏矩阵的相乘;

import java.util.*;

public class MultMat {

	public static void main(String[] args) {
		int i,j,k,l;
		Scanner scan = new Scanner(System.in);
		System.out.println("请输入矩阵M1的行数,列数,非零元的个数:");
		int mu1,nu1,tu1;
		mu1 = scan.nextInt();
		nu1 = scan.nextInt();
		tu1 = scan.nextInt();
		Mat M1 = new Mat(mu1,nu1,tu1);
		//输入矩阵M1
	    System.out.println("请输入矩阵M1的三元组:");
		for(i=1; i<=tu1; i++){
			M1.data[i].row = scan.nextInt();
			M1.data[i].col = scan.nextInt();
			M1.data[i].v = scan.nextInt();
		}
		System.out.println("输入的矩阵M1为:");
		M1.display();
		System.out.println("请输入矩阵M2的行数,列数,非零元的个数:");
		int mu2,nu2,tu2;
		mu2 = scan.nextInt();
		nu2= scan.nextInt();
		tu2 = scan.nextInt();
		Mat M2 = new Mat(mu2,nu2,tu2);
		//输入矩阵M2
		System.out.println("请输入矩阵M2的三元组:");
		for(i=1; i<=tu2; i++){
			M2.data[i].row = scan.nextInt();
			M2.data[i].col = scan.nextInt();
			M2.data[i].v = scan.nextInt();
		}
		System.out.println("输入的矩阵M2为:");
		M2.display();
		Mat M3 = new Mat(mu1,nu2,10);
		int ctemp[] = new int[mu1 + 1];
		int arow,brow=0,tp,t;//M1,M2的行数计数器
		
		/*M1.rpos[1] = 1;
		M1.rpos[2] = 3;
		M1.rpos[3] = 4;
		M2.rpos[1] = 1;
		M2.rpos[2] = 2;
		M2.rpos[3] = 3;
		M2.rpos[4] = 5;*/
		
		int num1[] = new int[mu1+10];
		int col1,row1,t1;
		for(col1=1; col1<=mu1; col1++){
			num1[col1] = 0;
		}
		for(t1=1; t1<=mu1; t1++){
			num1[M1.data[t1].row]++;
		}
		for(i=1; i<mu1; i++){
			M1.rpos[i] = M1.rpos[i - 1] + num1[i - 1];
		}
		
		int num2[] = new int[mu2+10];
		int col2,row2,t2;
		for(col2=1; col2<=mu2; col2++){
			num2[col2] = 0;
		}
		for(t2=1; t2<=mu1; t2++){
			num2[M2.data[t2].row]++;
		}
		for(i=1; i<mu1; i++){
			M2.rpos[i] = M2.rpos[i - 1] + num2[i - 1];
		}
		
		int p,q,ccol;
		for(arow=1; arow<=mu1; arow++){
			ctemp[arow] = 0;//当前各行元素累加器清零
			M3.rpos[arow] = M3.tu + 1;
			if(arow < mu1){
				tp = M1.rpos[arow + 1];
			}
			else{
				tp = mu1 + 1;
			}
			for(p=M1.rpos[arow]; p<tp; p++){
				brow = M1.data[p].col;
				//System.out.println(p + " " + brow);
				if(brow < mu2){
					t = M2.rpos[brow + 1];
				}
				else{
					t = tu2 + 1;
				}
				for(q=M2.rpos[brow]; q<t; q++){
					ccol = M2.data[q].col;
					ctemp[ccol] += M1.data[p].v * M2.data[q].v;
				}
			}
			for(ccol=1; ccol<=M3.nu; ccol++){
				if(ctemp[ccol]!=0){
					M3.data[M3.tu].row = arow;
					M3.data[M3.tu].col = ccol;
					M3.data[M3.tu].v = ctemp[ccol];
				}
			}
		 
		}
		
		System.out.println("相乘后的矩阵为:");
		M3.display();
		
	}

}


 

posted @ 2014-10-16 22:11  Pickle  阅读(736)  评论(0编辑  收藏  举报