確定罪犯藏身地點(diǎn)問題-資料下載頁
2025-06-07 21:26本頁面
【正文】 e(t,1)for j = 1:13y(i,j) = sqrt((t(i,1) a(j,1))*(t(i,1) a(j,1)) + (t(i,2) a(j,2))*(t(i,2) a(j,2)))。endendfor k=1:size(t,1)for l=1:13s(1,k)= y(k,l) + s(1,k)。endends = (s.*(1))./10000000。for n = 1: size(t,1)r(1,n) = exp(s(1,n))。endx = max(r)。h = find(r == x)w = r(1,h)w = h = 62t(62,:)ans =+006 * 新建m文件，名為diqitidiqiti(a,b,t(62,:))lest = []ib = 0。for d = (21:30)*500ib = 1 + iba = [[ , ][ , ][ , ][ , ][ , ][ , ][ , ][ , ][ , ][ , ][ , ][ , ][ , ]]t = []。for i=1:13e=[(a(i,1)+d),a(i,2)。(a(i,1)+2*d), a(i,2)。(a(i,1)d),a(i,2)。(a(i,1)2*d),a(i,2)。a(i,1),(a(i,2)+d)。a(i,1),(a(i,2)+2*d)。a(i,1),(a(i,2)d)。a(i,1),(a(i,2)2*d)。(a(i,1)+d),(a(i , 2)+d)。(a(i,1)+2*d),(a(i , 2)+2*d)。(a(i,1)d),(a(i , 2)d)。(a(i,1)2*d),(a(i , 2)2*d)。(a(i , 1)+d),(a(i , 2)d)(a(i , 1)+2*d),(a(i , 2)2*d)(a(i , 1)d),(a(i , 2)+d)。(a(i , 1)2*d),(a(i , 2)+2*d)]。for k=1:16for j= 1: 13if j~=iw(j,1) = sqrt((e(k,1) a(j,1))*(e(k,1) a(j,1)) + (e(k,2) a(j,2))*(e(k,2) a(j,2)))。endendif min(w)dh=e(k,:)t = [t。h]endendendty = []r = []s = zeros(1,size(t,1))。for i = 1:size(t,1)for j = 1:13y(i,j) = sqrt((t(i,1) a(j,1))*(t(i,1) a(j,1)) + (t(i,2) a(j,2))*(t(i,2) a(j,2)))。endendfor k=1:size(t,1)for l=1:13s(1,k)= y(k,l) + s(1,k)。endends = (s.*(1))./10000000。for n = 1: size(t,1)r(1,n) = exp(s(1,n))。endx = max(r)。h = find(r == x)w = r(1,h)。t(h,:)iblast(ib,1) = t(h,1)last(ib,2) = t(h,2)endlast結(jié)果為：last = +006 * 新建M文件 dibatifunction we = dibati(a,b,e)c = size(a,1)。r = 6400000。for i=1:c d(i,1) = r*cos(a(i))*cos(b(i))。 d(i,2) = r*cos(a(i))*sin(b(i))。endx1=(min(d(:,1))1250): 1250:(max(d(:,1))+1250)。y1=(min(d(:,2))2500): 2500:(max(d(:,2))+2500)。[X1,Y1] = meshgrid(x1,y1)。hold on。plot(X1,Y1,39。b39。)。plot(X139。,Y139。,39。b39。)。plot(d(:,1),d(:,2),39。o39。)for h = 1:10plot(e(h,1),e(h,2),39。.r39。)endhold off。vpa(d)運(yùn)行dibati(a,b,last)
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