数据结构-矩阵的压缩存储程序

实验报告课程名：数据结构（c 语言版） 实验名：矩阵的压缩存储姓名：班级：学号：时间： 2014.11.23精品资料一 实验目的与要求1. 掌握并实现稀疏矩阵的压缩存储的方法2. 在该存储方法上实现矩阵的操作二 实验内容? 判断一个用二维数组存储的矩阵是不是稀疏矩阵? 将其转化为压缩存储的形式? 在压缩存储上实现矩阵的乘法和转置操作三 实验结果与分析压缩转置程序：#include/判断该矩阵是否为稀疏矩阵#define m10#define n10 int amn=1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7,0,0,0,0,0,0,0,8,0,0,0,0,0,0,0,0,0,0,0,0,0,;struct threeint i,j;int value;struct three stu100;struct three1int i,j;int value;struct three1 stu1100;int jiance()int x=0;/ 赋初值为 0 for(x=0;x=99;x+)stux.value=0;float t=0; float v;for(int i=0;im;i+)for(int j=0;jn;j+)if(aij!=0)t+;if(v=t/(m*n)=0.05)printf( 该矩阵为稀疏矩阵 %fn,v); return 1;elseprintf( 该矩阵不是稀疏矩阵 n); return 0;void yasuo()int t=0;for(int r=0;rm;r+)for(int c=0;cn;c+)if(arc!=0)stut.i=r;stut.j=c; stut.value=arc; t+;void display()int x=0;printf( 压缩矩阵的三元组为 :n); for(x=0;x=99;x+)if(stux.value=0)break;printf(%d,%d,%d ,stux.i,stux.j,stux.value);printf(n);void zhuanzhi()int x=0;/ 赋初值为 0 int t=0;int num10=0,0,0,0,0,0,0,0,0,0;/每一列非 0 的数目for(x=0;x=99;x+)stu1x.value=0;for(int j=0;jn;j+)for(int i=0;im;i+)if(aij!=0)numj+; t+;int cpot10=0,0,0,0,0,0,0,0,0,0;cpot0=0; for(j=1;jn;j+)cpotj=cpotj-1+numj-1;int col=0; int q=0;for(int k=0;kt;k+)col=stuk.j; q=cpotcol;stu1q.i=stuk.j;stu1q.j=stuk.i; stu1q.value=stuk.value;+cpotcol;void display1()int x=0;printf( 转置以后的三元组为 :n); for(x=0;x=99;x+)if(stu1x.value=0) break;printf(%d,%d,%d ,stu1x.i,stu1x.j,stu1x.value);printf(n);void display2()int d,b; for(d=0;dm;d+)for(b=0;bm;b+)printf(%d,adb);printf(n);void main()display2(); if(jiance()=1)yasuo(); display(); zhuanzhi(); display1();图 1：压缩转置程序运行结果矩阵的乘法程序： #include #define m1 3#define n1 4#define m2 4#define n2 2int a1m1n1=3,0,0,5,0,-1,0,0,2,0,0,0,;int a2m2n2=0,2,1,0,-2,4,0,0,;struct three1int i,j;int value;struct three1 stu1100;struct three2int i,j;int value;struct three2 stu2100;struct three3int i,j;int value;struct three3 stu3100; int ar1posm1=0;int ar2posm2=0; int qrposm1;int yasuo1()int t=0; ar1pos0=0;for(int r=0;rm1;r+)for(int c=0;cn1;c+)if(a1rc!=0)stu1t.i=r;stu1t.j=c; stu1t.value=a1rc; t+;ar1posr+1=t;return t;int yasuo2()int t=0; ar2pos0=0;for(int r=0;rm2;r+)for(int c=0;cn2;c+)if(a2rc!=0)stu2t.i=r;stu2t.j=c; stu2t.value=a2rc; t+;ar2posr+1=t;return t;void chengfa(int x1,int x2)int a1m=0; int a2m=0;int tp,p,br,t,q,ccol;int qtu=0; for(a1m=0;a1mm1;a1m+)int ctempm1=0; if(a1m(m1-1)tp=ar1posa1m+1; elsetp=x1; for(p=ar1posa1m;ptp;+p)br=stu1p.j;if(br(m2-1)t=ar2posbr+1;elset=x2; for(q=ar2posbr;qt;+q)ccol=stu2q.j; ctempccol+=stu1p.value*stu2q.value;for(ccol=0;ccoln2;+ccol) if(ctempccol!=0)stu3qtu.i=a1m; stu3qtu.j=ccol; stu3qtu.value=ctempccol;+qtu;void display()int x=0;printf(a1与 a2 乘积之后的三元组是： n); for(x=0;x=99;x+)if(stu3x.value=0) break;printf(%d,%d,%d ,stu3x.i,stu3x.j,stu3x.value);printf(n);void display1()int m,n;printf(a1矩阵为： n); for(m=0;mm1;m+)for(n=0;nn1;n+)printf(%d,a1mn);printf(n);printf(a2矩阵为： n); for(m=0;mm2;m+)for(n=0;nn2;n+)printf(%d,a2m