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三元组顺序结构实现稀疏矩阵相加,行序优先(Java语言描述)

2014-10-16 23:40 961 查看
不用十字链表也可以稀疏矩阵相加时间复杂度最坏情况达到O(tuA + tuB);思路比较简单就不赘述了,代码如下:

三元组:

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;
}
}


构建矩阵存储结构:

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();
}
}
}

相加:

package 行逻辑链接的顺序表实现稀疏矩阵的相乘;

import java.util.*;

public class AddMat {

public static void main(String[] args) {
/*
* @author 王旭
* @time 2014/10/14/23:50
*
*/
Scanner scan =  new Scanner(System.in);
int muA,nuA,tuA,muB,nuB,tuB;
System.out.println("请输入A矩阵的行,列,非零元的个数:");
muA = scan.nextInt();
nuA = scan.nextInt();
tuA = scan.nextInt();
Mat A = new Mat(muA,nuA,tuA);
System.out.println("请输入A矩阵的三元组:");
for(int i=1; i<=tuA; i++){
A.data[i].row = scan.nextInt();
A.data[i].col = scan.nextInt();
A.data[i].v = scan.nextInt();
}
System.out.println("A矩阵为:");
A.display();
System.out.println("请输入B矩阵的行,列,非零元的个数:");
muB = scan.nextInt();
nuB = scan.nextInt();
tuB = scan.nextInt();
Mat B = new Mat(muB,nuB,tuB);
System.out.println("请输入B矩阵的三元组:");
for(int i=1; i<=tuB; i++){
B.data[i].row = scan.nextInt();
B.data[i].col = scan.nextInt();
B.data[i].v = scan.nextInt();
}
System.out.println("B矩阵为:");
B.display();
Mat C = new Mat(muA,nuA,1);
add(A,B,C);
System.out.println("相加后的矩阵C为:");
C.display();

}
public static void add(Mat A, Mat B, Mat C){
int k,l,temp;
//C = new Mat(A.mu,A.nu,1);
k = 1;
l = 1;
while(k<A.tu && l<B.tu){
if((A.data[k].row == B.data[l].row) && (A.data[k].col == B.data[l].col)){
temp = A.data[k].v + B.data[l].v;
if(temp != 0){
C.data[C.tu].row = A.data[k].row;
C.data[C.tu].col = A.data[k].col;
C.data[C.tu].v = temp;
C.tu++;
}
k++;
l++;
}
if(( (A.data[k].row == B.data[l].row) && (A.data[k].col < B.data[l].col) ) || (A.data[k].row<B.data[l].row)){
C.data[C.tu].row = A.data[k].row;
C.data[C.tu].col = A.data[k].col;
C.data[C.tu].v = A.data[k].v;
C.tu++;
k++;
}
if(( (B.data[l].row == A.data[k].row) && (B.data[l].col < A.data[k].col) ) || (B.data[l].row<A.data[k].row)){
C.data[C.tu].row = B.data[l].row;
C.data[C.tu].col = B.data[l].col;
C.data[C.tu].v = B.data[l].v;
C.tu++;
l++;
}
}
}

}
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