sourcePic=imread（‘xcv-2.jpg‘）;
sourcePic=rgb2gray（sourcePic）;

sourcePic=im2double（sourcePic）;

%使用水平和垂直Sobel算子自動(dòng)選擇閾值?
edgePic=edge（sourcePic‘sobel‘‘horizontal‘）;?
figureimshow（edgePic）title（‘水平圖像邊緣檢測(cè)‘）;?%?顯示邊緣探測(cè)圖像?

edgePic=edge（sourcePic‘sobel‘‘vertical‘）;?

figureimshow（edgePic）title（‘垂直圖像邊緣檢測(cè)‘）;?%?顯示邊緣探測(cè)圖像?

edgePic=edge（sourcePic‘sobel‘）;?

[heightwidth]=size（edgePic）;?

?%圖像中直線的個(gè)數(shù)
?LINE_COUNT?=?100;
?
?%?Hough變換檢測(cè)直線，g（x）=（ap）為邊界點(diǎn)對(duì)應(yīng)的平面??
?ma=180;?%a的值為0到180度??

?mp=round（sqrt（height^2+width^2））;?%對(duì)應(yīng)P的最大值??

?npc=zeros（ma2*mp）;?%用于記錄（ap）對(duì)應(yīng)的點(diǎn)的個(gè)數(shù)??
?
?MAP=zeros（height?width）;
?
?npp=cell（ma2*mp）;?%用于記錄（ap）對(duì)應(yīng)的點(diǎn)的坐標(biāo)??
?
?for?i=1:height?%計(jì)算（ap）的值，并做相應(yīng)記錄??

?????for?j=1:width??

?????????if（edgePic（ij）==1）??

?????????????for?k=1:ma
?????????????????p?=?round（i*cos（pi*k/180）+j*sin（pi*k/180））;
?????????????????npc（kmp+p）=npc（kmp+p）+1;
?????????????????npp{kmp+p}=[npp{kmp+p}[ij]‘];
?????????????end??

?????????end??

?????end??

?end


LNum?=?0;???????????????????????%得到的直線的數(shù)目
LPC?=?zeros（LINE_COUNT6）;??????%直線的參數(shù)集合?元素為?[Number][k][b][a][p][DP]???分別為直線的:?編號(hào)?斜率?截距?角度a?弦長(zhǎng)p?是否為消失線
index?=?1;


%提取直線的閥值
LT?=?90;

for?i=1:ma??????

?????for?j=1:mp*2??

?????????if（npc（ij）?>?LT）?
?????????????
????????????????lk=?-cot（pi*i/180）;
????????????????lb=?round（（j?-?mp?-?1）/sin（pi*i/180））;

????????????????%保存直線的參數(shù)，順序?yàn)?編號(hào)?（kb）?（ap）
????????????????LNum?=?LNum?+?1;
????????????????
????????????????LPC（index1）?=?LNum;%存儲(chǔ)直線的編號(hào)
????????????????LPC（index2）?=?lk;
????????????????LPC（index3）?=?lb;
????????????????LPC（index4）?=?i;
????????????????LPC（index5）?=?j;
????????????????
????????????????index?=?index?+?1;
????????????????
????????????????lp=npp{ij};
????????????????for?k=1:npc（ij）
?????????????????????MAP（lp（1k）lp（2k））=1;??
????????????????end
?????????end??

?????end??

end??

figureimshow（MAP）;title（‘檢測(cè)出的直線圖‘）;

%求出總共有多少交點(diǎn)
InterCount?=?LNum*（LNum?-?1）*0.5;

LInter?=?zeros（InterCount4）;???????????%直線的交點(diǎn)集合?元素為?兩條相交直線的編號(hào)?ij?和?交點(diǎn)坐標(biāo)?xy?.?有?i?????????????????????????????????????????%元素個(gè)數(shù)為?n*（n?-?1）/2
%下面的代碼是求直線的交點(diǎn)
T?=1;

LIC?=?0;????????????????????????????????%直線交點(diǎn)的個(gè)數(shù)

for?i?=?1:LNum?-?1
????for?j?=?i?+?1:LNum
????????
????????k1?=?LPC（i2）;
????????b1?=?LPC（i3）;

????????k2?=?LPC（j2）;
????????b2?=?LPC（j3）;
????????
????????dp?=?k1?-?k2;???????????????????%得到兩條直線斜率之差
????????
????????if（abs（dp?-?0）?>?10e-03）????????%如果斜率之差太小，就可以認(rèn)為這兩條直線平行而沒(méi)有交點(diǎn)。
????????????
????????????LIC?=?LIC?+?1;??????????????%直線交點(diǎn)個(gè)數(shù)加一
????????????
????????????X?=?（b2?-?b1）/（k1?-?k2）;????%?X?=?（b2?-?b1）/（k1?-?k2）
????????????Y?=?k1?*?X?+?b1;????????????%?Y?=?k1?*?X?+?b1
????????????LInter（LIC1）?=?i;
????????????LInter（LIC2）?=?j;
????????????LInter（LIC3）?=?round（X）;
????????????LInter（LIC4）?=?round（Y）;
????????????
????????end
????end
end


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%以下的部分是求消失點(diǎn)和消失線的參數(shù)
%%%

?屬性????????????大小?????日期????時(shí)間???名稱(chēng)
-----------?---------??----------?-----??----
?????文件????????9092??2012-09-04?18:13??All.m