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

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <queue>
 7 #include <climits>
 8 #include <cmath>
 9 #include <cstdlib>
10
11 using namespace std;
12
13 #define ll long long
14 #define N 1000100
15 char a[N] , b[N];
16 int _next[N] , extend[N] , len1 , len2;
17
18 void get_next(char *T){// _next[i]: 以第i位置开始的子串 与 T的公共前缀
19      int i,length = strlen(T);
20      _next[0] = length;
21      for(i = 0;i<length-1 && T[i]==T[i+1]; i++);
22           _next[1] = i;
23           int a = 1;
24           for(int k = 2; k < length; k++){
25                   int p = a+_next[a]-1, L = _next[k-a];
26                   if( (k-1)+L >= p ){
27                        int j = (p-k+1)>0? (p-k+1) : 0;
28                        while(k+j<length && T[k+j]==T[j]) j++;// 枚举(p+1，length) 与(p-k+1,length) 区间比较
29                        _next[k] = j, a = k;
30                   }
31                   else _next[k] = L;
32          }
33 }
34 void getextend(char *S,char *T){
35     int maxl=max(len1 , len2);
36     for(int i=0 ; i<=maxl ; i++) _next[i]=0;
37     get_next(T);
38     int Slen = strlen(S), Tlen = strlen(T), a = 0;
39     int MinLen = Slen>Tlen?Tlen:Slen;
40     while(a<MinLen && S[a]==T[a]) a++;
41     extend[0] = a, a = 0;
42     for(int k = 1; k < Slen; k++){
43       int p = a+extend[a]-1, L = _next[k-a];
44       if( (k-1)+L >= p ){
45            int j = (p-k+1)>0? (p-k+1) : 0;
46            while(k+j<Slen && j<Tlen && S[k+j]==T[j] ) j++;
47            extend[k] = j;a = k;
48       }
49       else extend[k] = L;
50     }
51 }
52
53
54 int main()
55 {
56     #ifndef ONLINE_JUDGE
57         freopen("a.in" , "r" , stdin);
58     #endif
59     while(~scanf("%s%s" , a , b))
60     {
61         getextend(b , a);
62         len1 = strlen(a) , len2 = strlen(b);
63         int ret = 0;
64         for(int i=0 ; i<len2 ; i++)
65             if(extend[i]>ret&&extend[i]+i==len2) ret = extend[i];
66         if(!ret) puts("0");
67         else{
68             strncpy(a , b+(len2-ret) , ret);
69             a[ret] = ‘\0‘;
70             printf("%s %d\n" , a, ret);
71         }
72     }
73     return 0;
74 }