HDU 2594 扩展kmp模板题

　题目大意：

给定两个字符串，在第一个字符串中找到一个最大前缀作为第二个字符串的后缀

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <climits>
 #include <cmath>
 #include <cstdlib>

 using namespace std;

 #define ll long long
 #define N 1000100
 char a[N] , b[N];
 int _next[N] , extend[N] , len1 , len2;

 void get_next(char *T){// _next[i]: 以第i位置开始的子串 与 T的公共前缀
      int i,length = strlen(T);
      _next[] = length;
      for(i = ;i<length- && T[i]==T[i+]; i++);
           _next[] = i;
           int a = ;
           for(int k = ; k < length; k++){
                   int p = a+_next[a]-, L = _next[k-a];
                   if( (k-)+L >= p ){
                        int j = (p-k+)>? (p-k+) : ;
                        while(k+j<length && T[k+j]==T[j]) j++;// 枚举(p+1，length) 与(p-k+1,length) 区间比较
                        _next[k] = j, a = k;
                   }
                   else _next[k] = L;
          }
 }
 void getextend(char *S,char *T){
     int maxl=max(len1 , len2);
     for(int i= ; i<=maxl ; i++) _next[i]=;
     get_next(T);
     int Slen = strlen(S), Tlen = strlen(T), a = ;
     int MinLen = Slen>Tlen?Tlen:Slen;
     while(a<MinLen && S[a]==T[a]) a++;
     extend[] = a, a = ;
     for(int k = ; k < Slen; k++){
       int p = a+extend[a]-, L = _next[k-a];
       if( (k-)+L >= p ){
            int j = (p-k+)>? (p-k+) : ;
            while(k+j<Slen && j<Tlen && S[k+j]==T[j] ) j++;
            extend[k] = j;a = k;
       }
       else extend[k] = L;
     }
 }

 int main()
 {
     #ifndef ONLINE_JUDGE
         freopen("a.in" , "r" , stdin);
     #endif
     while(~scanf("%s%s" , a , b))
     {
         getextend(b , a);
         len1 = strlen(a) , len2 = strlen(b);
         int ret = ;
         for(int i= ; i<len2 ; i++)
             if(extend[i]>ret&&extend[i]+i==len2) ret = extend[i];
         if(!ret) puts("");
         else{
             strncpy(a , b+(len2-ret) , ret);
             a[ret] = '\0';
             printf("%s %d\n" , a, ret);
         }
     }
     return ;
 }