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#include #include#include #include #include#include #include#include#include#includeusing namespace std; #define lch(i) ((i)<<1) #define rch(i) ((i)<<1|1) #define sqr(i) ((i)*(i))

注意边界#include #include#include #include #include#include #include#include#include#includeusing namespace std; #define lch(i) ((i)<<1) #define rch(i) ((i)<<1|1) #define sqr(i) ((i

这题只有像素变化的地方，该点包括周围八个点都有变化的可能。举个例子：。。。。。。。。。。 ，，，，，，，，，，= =  = =  =  =  =  =  =  =假设，的四周都没有像素变化的点，那么所有，的编码值都等于max(abs（，减。）,abs(，减=)引用别人一个清晰的代码传送门：http://blog.youkuaiyun.com/lyy289065406/article

#include #include#include #include #include#include #include#include#include#includeusing namespace std; #define lch(i) ((i)<<1) #define rch(i) ((i)<<1|1) #define sqr(i) ((i)*(i))

#include #include#include #include #include#include #include#include#include#includeusing namespace std; #define lch(i) ((i)<<1) #define rch(i) ((i)<<1|1) #define sqr(i) ((i)*(i))

#pragma warning(disable:4996)#include #include#include #include #include#include #include#include#include#includeusing namespace std; #define lch(i) ((i)<<1) #define rch(i) ((i)<<1

#include #include#include #include #include#include #include#include#include#includeusing namespace std; #define lch(i) ((i)<<1) #define rch(i) ((i)<<1|1) #define sqr(i) ((i)*(i))

#include<iostream> #include<string> #include<string.h>#include<algorithm> #include<fstream>#include<stack>#include<map>using namespace std; #define lch(i) ((i)<<1) #define rch(i) ((i)<<1|