matlab如何求一个N*2的矩阵的任意两点间的距离? MATLAB 求任意两点间的距离并生成矩阵

\u7528MATLAB\u6c42\u4efb\u610f\u4e24\u70b9\u95f4\u8ddd\u79bb

close all
clear,clc

A = [413359
403343
383.5351
381377.5
339376
335383
317362
334.5353.5
333342
282325
247301
219316
225270
280292
290335
337328
415335
432371
418374];

X = A(:,1);
Y = A(:,2);

N = length(X);
D = zeros(N,N);

for I = 2:N
for J = 1:I-1
D(I,J) = sqrt((X(I) - X(J))*(X(I) - X(J)) + (Y(I) - Y(J))*(Y(I) - Y(J)));
end
end

D % \u4efb\u610f\u4e24\u70b9\u95f4\u8ddd\u79bb
D1 = D+D' % \u4efb\u610f\u4e24\u70b9\u95f4\u8ddd\u79bb

\u4f60\u7684\u610f\u601d\u5e94\u8be5\u662f\u6784\u9020\u90bb\u63a5\u77e9\u9635\uff0c\u76f4\u63a5\u5bf9\u77e9\u9635\u4e2d\u7684\u6bcf\u4e2a\u5143\u7d20\u4f9d\u6b21\u8d4b\u503c\u5c31\u53ef\u4ee5\u3002

\u5177\u4f53\u5b9e\u73b0\u4e3b\u8981\u770b\u4f60\u4f7f\u7528\u7684\u662f\u4ec0\u4e48\u6570\u636e\u5b58\u50a8\u65b9\u5f0f\uff1a
\uff08\u4e00\u79cd\u65b9\u6cd5\uff0c\u4ec5\u4f9b\u53c2\u8003\uff09

\u5982\u679c\u4f60\u628a\u70b9\u50a8\u5b58\u6210\u4e00\u79cd\u7ed3\u6784\uff0c\u8bb0\u90bb\u63a5\u77e9\u9635\u4e3aA\uff0c\u5219A(i,j)=dist(i,j)\uff0cdist\u662f\u81ea\u5df1\u5b9a\u4e49\u7684ij\u4e24\u70b9\u4e4b\u95f4\u8ddd\u79bb\u8ba1\u7b97\u7684\u51fd\u6570


\u8fd8\u6709\u5f88\u591a\u5b9e\u73b0\u7684\u65b9\u6cd5\uff0c\u5bf9\u4e8e\u5c0f\u6570\u636e\u91cf\uff0c\u4e2a\u4eba\u89c9\u5f97\u90fd\u6ca1\u4ec0\u4e48\u5dee\u522b\u3002

你这是求任意点的最短距,直接用欧拉公式,或叫欧氏距离求法,
问题矩阵C中有31个点,这样最短距矩阵是d(31*31),求解如下:
%1、求任意点距离
for i=1:31
for j=1:31
C1(i,j)=sqrt((C(i,1)-C(j,1))^2+(C(i,2)-C(j,2))^2);
end
end
%2、因为每个点到自身的距离为0,故求L之前要做处理,将每个点到自身的距离设为无限大:
for i=1:31
C1(i,i)=inf;
end
%3、求最短距L
[L,mini]=min(C1(:))

解得;
任意两点距离
C1(太多了,粘贴后不能发表,请复制上述问题自己求解。)

最短距L=
111.2879
mini = 35(表明在C1矩阵中标号为第2点和第4点之间)

另外,如果求带路径问题,如真实的道路路口距离可以用floyd算法,
function [D,path,min1,path1]=floyd(a,start,terminal)
D=a;n=size(D,1);path=zeros(n,n);
for i=1:n
for j=1:n
if D(i,j)~=inf
path(i,j)=j;
end, end, end
for k=1:n
for i=1:n
for j=1:n
if D(i,k)+D(k,j)<D(i,j)
D(i,j)=D(i,k)+D(k,j);
path(i,j)=path(i,k);
end, end, end,end
if nargin==3
min1=D(start,terminal);
m(1)=start;
i=1;
path1=[ ];
while path(m(i),terminal)~=terminal
k=i+1;
m(k)=path(m(i),terminal);
i=i+1;
end
m(i+1)=terminal;
path1=m;
end

计算点之间的距离,距离矩阵distance(N*N),输出到表格中distance.xls。
%% Test_repmat
C=[1304,2312; 3639,1315; 4177,2244; 3712,1399; 3488,1535;3326,1556; 3238,1229;
4196,1004; 4312,790; 4386,570;3007,1970; 2562,1756; 2788,1491; 2381,1676;
1332, 695;3715,1678; 3918,2179; 4061,2370; 3780,2212; 3676,2578;4029,2838;
4263,2931; 3429,1908; 3507,2367; 3394,2643;3439,3201; 2935,3240; 3140,3550;
2545,2357; 2778,2826;2370,2975; ];
p = C;
k = size(p,1);
P_x = repmat(p(:,1),1,k);
P_y = repmat(p(:,2),1,k);
Pt_x = repmat(p(:,1)',k,1);
Pt_y = repmat(p(:,2)',k,1);
distance = sqrt((P_x - Pt_x).^2 + (P_y - Pt_y).^2);
xlswrite('distance.xls',distance)
%by [email protected]

建议你查一下Dijkstra算法和Floyd算法或者Kruskal或Prim算法,要想学好matlab还得多自学点东西~~~

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