力导向图force-direct算法（MATLAB实现）含数据集

DATA = int32(xlsread('bus685.xlsx',1,'A1:B1282'))+1; position =xlsread('position.xlsx',1,'A1:B685'); K_r = 0.1; K_s = 0.025; L = 2; delta_t = 10; MaxLength =8; tic for k = 1:100 force = zeros(685,2); for m = 2:684 for n = m+1:685 dx = position(n,1)-position(m,1); dy = position(n,2)-position(m,2); if dx~=0 || dy~=0 dist2 = dx^2+dy^2; dist = sqrt(dist2); f = K_r/dist2; fx = f*dx/dist; fy = f*dy/dist; force(m,1) = force(m,1)-fx; force(m,2) = force(m,2)-fy; force(n,1) = force(n,1)+fx; force(n,2) = force(n,2)+fy; end end end for m = 2:685 ne=find(DATA(:,1)==m); for n = 1:length(ne) dx = position(DATA(ne(n),2),1)-position(m,1); dy = position(DATA(ne(n),2),2)-position(m,2); if dx~=0||dy~=0 dist = sqrt(dx^2+dy^2); f = K_s*(dist-L); fx = f*dx/dist; fy = f*dy/dist; force(m,1) = force(m,1)+fx; force(m,2) = force(m,2)+fy; force(DATA(ne(n),2),1) = force(DATA(ne(n),2),1)-fx; force(DATA(ne(n),2),2) = force(DATA(ne(n),2),2)-fy; end end end for i = 2:685 dx = delta_t*force(i,1); dy = delta_t*force(i,2); displacement = dx^2+dy^2; if(displacement>MaxLength) s = sqrt(MaxLength/displacement); dx = s*dx; dy = s*dy; end position(i,1) = position(i,1)+dx; position(i,2) = position(i,2)+dy; end end toc for i = 1:1282 s = DATA(i,1); d = DATA(i,2); x = [position(s,1);position(d,1)]; y = [position(s,2);position(d,2)]; plot(x,y,'o-b'); hold on; end sum = 0; for j = 1:685 sum = sum+abs(force(j,1))+abs(force(j,2)); end sum