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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%ncc算法%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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function?res=match（a1cnt1r1c1a2cnt2r2c2）
%?res=match（a1a2）
%?將從a1尋找a2中的最佳匹配點，得到從a2中抽取的res，也就是單向搜索
%?[result1cnt1r11c11]=harris（a1）;
%?[result2cnt2r22c22]=harris（a2）;%可以保證想匹配哪些點就匹配哪些
%?figure;
%??imshow（result1）;title（‘result1角點位置‘）;
%??figure;title（‘result2角點位置‘）;
%??imshow（result2）;???
?
win=[1/9?1/9?1/9;1/9?1/9?1/9;1/9?1/9?1/9];
u1=filter2（wina1）;
u2=filter2（wina2）;??%求均值
??????????????????????????
??????

a1=double（a1）;
a2=double（a2）;
A=filter2（win（a1-u1）.^2）;%求方差
B=filter2（win（a2-u2）.^2）;
[m1n1]=size（a1）;
[m2n2]=size（a2）;
res1=zeros（m1n1）;
res2=zeros（m2n2）;???????????%尋找的匹配的點

for?s=1:cnt1???????????
????????max=0;?p=0;q=0;i=r1（s1）;j=c1（s1）;??????%p.q存放坐標(biāo)
????????for?v=1:cnt2
????????????m=r2（v1）;n=c2（v1）;
????????????u1（ij）=（a1（i-1j-1）+a1（i-1j）+a1（i-1j+1）+a1（ij-1）+a1（ij）+a1（ij+1）+a1（i+1j-1）+a1（i+1j）+a1（i+1j+1））/9;%%求均值
????????????u2（mn）=（a2（m-1n-1）+a2（m-1n）+a2（m-1