空間網(wǎng)格節(jié)點(diǎn)最小距離

1、1 已知有一xyz坐標(biāo)系中的空間網(wǎng)格以及網(wǎng)格在xy面上的投影（參看下圖）。（1）試求點(diǎn)從節(jié)點(diǎn)A(0,0，z(0,0)沿著空間網(wǎng)格邊走到B(1,1,z(1,1)的最短路線的距離；（2）如果允許沿著對(duì)角線行走，求上述的最短路線的距離；（3）如果行走路線要求過村莊C(0.6,00.3),請求最短路線的距離。（1）問題1程序如下sets:gg/1.6/;link1(gg,gg):z;link2(link1,link1):w,x;endsetsdata:w=1000;z= 10.0000 9.9600 9.8400 9.6400 9.3600 9.0000 9.9600 9.9200 9.8000 9.

2、6000 9.3200 8.9600 9.8400 9.8000 9.6800 9.4800 9.2000 8.8400 9.6400 9.6000 9.4800 9.2800 9.0000 8.6400 9.3600 9.3200 9.2000 9.0000 8.7200 8.3600 9.0000 8.9600 8.8400 8.6400 8.3600 8.0000;enddatacalc:for(link2(i,j,m,n)|i#lt#6 #and# m#eq#i+1 #and# n#eq#j:w(i,j,m,n)=sqrt(2*0.22+(z(i,j)-z(m,n)2);for(lin

3、k2(i,j,m,n)|j#lt#6 #and# m#eq#i #and# n#eq#j+1:w(i,j,m,n)=sqrt(2*0.22+(z(i,j)-z(m,n)2);for(link2(i,j,m,n):w(m,n,i,j)=w(i,j,m,n);endcalcmin=sum(link2(i,j,m,n):w(i,j,m,n)*x(i,j,m,n);for(link1(i,j)|(i+j)#ne#2 #and# (i+j)#ne#12:sum(link1(s,t):x(i,j,s,t)=sum(link1(s,t):x(s,t,i,j);sum(link1(m,n):x(1,1,m,n

4、)=1;sum(link1(i,j):x(i,j,1,1)=0;sum(link1(i,j):x(i,j,6,6)=1;sum(link1(m,n):x(6,6,m,n)=0;for(link2:bin(x);end（2）問題2程序如下（只需添加一個(gè)語句命令）sets:gg/1.6/;link1(gg,gg):z;link2(link1,link1):w,x;endsetsdata:w=1000;z= 10.0000 9.9600 9.8400 9.6400 9.3600 9.0000 9.9600 9.9200 9.8000 9.6000 9.3200 8.9600 9.8400 9.800

5、0 9.6800 9.4800 9.2000 8.8400 9.6400 9.6000 9.4800 9.2800 9.0000 8.6400 9.3600 9.3200 9.2000 9.0000 8.7200 8.3600 9.0000 8.9600 8.8400 8.6400 8.3600 8.0000;enddatacalc:for(link2(i,j,m,n)|i#lt#6 #and# m#eq#i+1 #and# n#eq#j:w(i,j,m,n)=sqrt(2*0.22+(z(i,j)-z(m,n)2);for(link2(i,j,m,n)|j#lt#6 #and# m#eq#i

6、 #and# n#eq#j+1:w(i,j,m,n)=sqrt(2*0.22+(z(i,j)-z(m,n)2);for(link2(i,j,m,n)|i#lt#6#and#j#lt#6 #and# m#eq#i+1 #and# n#eq#j+1:w(i,j,m,n)=sqrt(2*0.22+(z(i,j)-z(m,n)2);for(link2(i,j,m,n):w(m,n,i,j)=w(i,j,m,n);endcalcmin=sum(link2(i,j,m,n):w(i,j,m,n)*x(i,j,m,n);for(link1(i,j)|(i+j)#ne#2 #and# (i+j)#ne#12:

7、sum(link1(s,t):x(i,j,s,t)=sum(link1(s,t):x(s,t,i,j);sum(link1(m,n):x(1,1,m,n)=1;sum(link1(i,j):x(i,j,1,1)=0;sum(link1(i,j):x(i,j,6,6)=1;sum(link1(m,n):x(6,6,m,n)=0;for(link2:bin(x);End（3）問題3程序如下（可以先進(jìn)行插值，把控制點(diǎn)作為前一段的終點(diǎn)，在作為后一段的始點(diǎn)，分兩次與前面同樣的方法進(jìn)行計(jì)算）（4）說明：當(dāng)目標(biāo)點(diǎn)不是右上角頂點(diǎn)時(shí)處理如下sets:gg/1.6/;link1(gg,gg):z;link2(li

8、nk1,link1):w,x;endsetsdata:w=1000;z= 10.0000 9.9600 9.8400 9.6400 9.3600 9.0000 9.9600 9.9200 9.8000 9.6000 9.3200 8.9600 9.8400 9.8000 9.6800 9.4800 9.2000 8.8400 9.6400 9.6000 9.4800 9.2800 9.0000 8.6400 9.3600 9.3200 9.2000 9.0000 8.7200 8.3600 9.0000 8.9600 8.8400 8.6400 8.3600 8.0000;enddatacal

9、c:for(link2(i,j,m,n)|i#lt#6 #and# m#eq#i+1 #and# n#eq#j:w(i,j,m,n)=sqrt(0.22+(z(i,j)-z(m,n)2);for(link2(i,j,m,n)|j#lt#6 #and# m#eq#i #and# n#eq#j+1:w(i,j,m,n)=sqrt(0.22+(z(i,j)-z(m,n)2);for(link2(i,j,m,n)|i#lt#6 #and# j#lt#6 #and# m#eq#(i+1) #and# n#eq#(j+1):w(i,j,m,n)=sqrt(2*0.22+(z(i,j)-z(m,n)2);f

10、or(link2(i,j,m,n):w(m,n,i,j)=w(i,j,m,n);endcalcmin=sum(link2(i,j,m,n):w(i,j,m,n)*x(i,j,m,n);for(link1(i,j)|(i+j)#ne#2 #and# i#ne#6 #and# j#ne#6:sum(link1(s,t):x(i,j,s,t)=sum(link1(s,t):x(s,t,i,j);for(link1(i,j)|i #eq#6 #and# j#ne#5:sum(link1(s,t):x(i,j,s,t)=sum(link1(s,t):x(s,t,i,j);for(link1(i,j)|j #eq#5 #and# i#ne#6:sum(link1(s,t):x(i,j,s,t)=sum(link1(s,t):x(s,t,i,j);sum(link1(m,n):x(1,1,m,n)=