Time Limit: 24000/12000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 981 Accepted Submission(s): 471
Two chain codes may represent the same shape if the shape has been rotated, or if a different starting point is chosen for the contour. To normalize the code for rotation, we can compute the first difference of the chain code instead. The first difference is obtained by counting the number of direction changes in counterclockwise direction between consecutive elements in the chain code (the last element is consecutive with the first one). In the above code, the first difference is
00110026202011676122
Finally, to normalize for the starting point, we consider all cyclic rotations of the first difference and choose among them the lexicographically smallest such code. The resulting code is called the shape number.
00110026202011676122
01100262020116761220
11002620201167612200
...
20011002620201167612
In this case, 00110026202011676122 is the shape number of the shape above.
#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 300005 const int inf=0x7fffffff; //无限大 string s; char b[maxn]; int getminsub(char *a) { int i=0,j=1,len=strlen(a),k=0; //取两个同构的字符串一个从下标0开始,一个从下标1开始 while(i<len&&j<len&&k<len) //这里并没有将字符串复制一份添加到后面 { if(k==len) break; //说明找到了a的最小表示 if(i==j) j++; int ni=i+k,nj = j+k; if(ni>=len) ni-=len; //就是回到字符串的开始去 if(nj>=len) nj-=len; if(a[ni]>a[nj]) { i+=k+1; k=0; } else if(a[ni]<a[nj]) { j+=k+1; k=0; } else k++; } return i; //返回从第i个字符开始时a的最小表示 } int main() { while(cin>>s) { int n=s.size(); for(int i=0;i<s.size();i++) { if(s[(i+1)%n]>=s[i]) b[i]=(s[(i+1)%n]-s[i])+'0'; else b[i]=s[(i+1)%n]+8-s[i]+'0'; } int k=getminsub(b); for(int i=k;i<s.size()+k;i++) { cout<<b[i%n]; } cout<<endl; } }