2013建模竞赛B题讲稿.docx

2013全国大学生数学建模竞赛B题碎纸片的拼接复原%%%%%%%%%%%%%%%%%%%%%%%%%第一问的算法流程%%%%%%%%%%%%%a{1}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\0.bmp);a{2}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\1.bmp);a{3}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\2.bmp);a{4}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\3.bmp);a{5}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\4.bmp);a{6}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\5.bmp);a{7}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\6.bmp);a{8}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\7.bmp);a{9}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\8.bmp);a{10}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\9.bmp);a{11}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\10.bmp);a{12}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\11.bmp);a{13}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\12.bmp);a{14}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\13.bmp);a{15}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\14.bmp);a{16}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\15.bmp);a{17}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\16.bmp);a{18}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\17.bmp);a{19}=imread(C:\Users\Aero?Fantasia\Desktop\Competition\B1\T1\18.bmp);由于该像素图片转换后为的矩阵，论文中无法放置，所以仅简单举例说明，矩阵的中元素表示该位置图片的灰度，255表示为白，0为黑，图片中信息为黑白文字信息，但由于文字信息会存在阴影，所以矩阵中出现了介于0-255的元素。每一个图片转换后都是一个的矩阵，矩阵中的元素的值介于0-255之间。然后利用下面的程序取第一列和最后一列得到矩阵T{i}，是一个的矩阵，第一列表示第i个碎片的左边灰度值，第二列表示第i个碎片的右边灰度值。%取边缘for?i=1:19???demo=[];???demo(:,1)=a{i}(:,1);???demo(:,2)=a{i}(:,length(a{i}(1,:)));???T{i}=demo;end然后将灰度值矩阵转化为0-1值矩阵%二值化for?i=1:19????Com=[];???for?m=1:2??????for?n=1:1980??????????if?T{i}(n,m)==255??????????????Com(n,m)=0;??????????else??????????????Com(n,m)=1;??????????end??????end???end???C{i}=Com;end例如可将转化为如下的矩阵：其中白色用0值表示，非白色用1表示。步骤二：基于上述 0-1 矩阵，提取每幅图片左右边界的 0-1 值，建立边缘匹配度模型，确定出图片的序列；先观察0-1矩阵，发现有两个矩阵