数学建模之输油管的布置

double h,y2,x2,w;

double a,b,c; h=0.13; y2=0; x2=5; int i,j,k; double min=10000; for(i=0;i<=20;i++) {h=h+0. 1; y2=6;

for(j=0;j<=20;j++)

{y2=y2+0. 1; x2=12;

for(k=0;k<=20;k++) {x2=x2+0.1;

w=27.6*sqrt((8-y2)*(8-y2)+25)+5.6*sqrt((5-h)*(5-h)+(20-x2)*(20-x2))+6*sqrt((x2-5)*(x2-5)+(y2-h)*(y2-h))+7.2*h;

if(min>w) {

min=w;

a=h;

循环最终可得到