如何獲得線矩形交叉段？

Question

我想找到代數重建方法的權重矩陣。 為此，我必須找到與網格的線交點。 我可以找到直線與線的交點，但我必須存儲相交的線段網格數。 因此，假設在網格第一個正方形中不與網格相交，則將零置於權重矩陣的第一個元素上。

這是我為線路交叉嘗試的代碼：

ak = 3:6
aka = 3:6
x = zeros(size(aka))
y = zeros(size(ak))
for k = 1:length(ak)
  line([ak(1) ak(end)], [aka(k) aka(k)],'color','r')
end

% Vertical grid
for k = 1:length(aka)
  line([ak(k) ak(k)], [aka(1) aka(end)],'color','r')
end
hold on;
 X =[0 15.5]
 Y = [2.5 8.5] 
 m = (Y(2)-Y(1))/(X(2)-X(1)) ;
 c = 2.5 ; 
 plot(X,Y)
axis([0 10 0 10])
axis square
% plotting y intercept
for i = 1:4
    y(i) = m * ak(i) + c
    if y(i)<2 || y(i)>6
        y(i) = 0
    end
end
% plotting x intercept
for i = 1:4
   x(i) = (y(i) - c)/m 
    if x(i)<2 || x(i)>6
        x(i) = 0
    end
end  
z = [x' y']

我有一條線，由參數m, h定義，其中y = m*x + h這條線穿過網格（即像素）。

對於網格的每個方格(a, b) （即方形[a, a+1]x[b, b+1] ），我想確定給定的線是否穿過這個方格 ，如果是， 廣場中段的長度是多少，這樣我就可以構造權重矩陣，這對於代數重建方法是必不可少的。

Answer 1

這是一個很好的方法，使一條線與矩形網格相交並得到每個交叉段的長度：我使用了來自此鏈接的第三個答案中的偽代碼的線條交點

% create some line form the equation y=mx+h
m = 0.5; h = 0.2;
x = -2:0.01:2;
y = m*x+h;
% create a grid on the range [-1,1]
[X,Y] = meshgrid(linspace(-1,1,10),linspace(-1,1,10));
% create a quad mesh on this range
fvc = surf2patch(X,Y,zeros(size(X)));
% extract topology
v = fvc.vertices(:,[1,2]);
f = fvc.faces;
% plot the grid and the line
patch(fvc,'EdgeColor','g','FaceColor','w'); hold on;
plot(x,y);
% use line line intersection from the link
DC = [f(:,[1,2]);f(:,[2,3]);f(:,[3,4]);f(:,[4,1])];
D = v(DC(:,1),:);
C = v(DC(:,2),:);
A = repmat([x(1),y(1)],size(DC,1),1);
B = repmat([x(end),y(end)],size(DC,1),1);
E = A-B;
F = D-C;
P = [-E(:,2),E(:,1)];
h = dot(A-C,P,2)./dot(F,P,2);
% calc intersections
idx = (0<=h & h<=1);
intersections = C(idx,:)+F(idx,:).*repmat(h(idx),1,2);
intersections = uniquetol(intersections,1e-8,'ByRows',true);
% sort by x axis values
[~,ii] = sort(intersections(:,1));
intersections = intersections(ii,:);
scatter(intersections(:,1),intersections(:,2));
% get segments lengths
directions = diff(intersections);
lengths = sqrt(sum(directions.^2,2));
directions = directions./repmat(sqrt(sum(directions.^2,2)),1,2);
directions = directions.*repmat(lengths,1,2);
quiver(intersections(1:end-1,1),intersections(1:end-1,2),directions(:,1),directions(:,2),'AutoScale','off','Color','k');

這是結果（圖像中箭頭的長度是段長度）