Jobdu-1535-重叠的最长子串[扩展KMP] | Acm之家


九度-1535-重叠的最长子串[扩展KMP] | Acm之家
给定两个字符串,求它们前后重叠的最长子串的长度,比如"abcde"和"cdefg"是"cde",长度为3。


对于每个测试案例只有一行, 包含两个字符串。字符串长度不超过1000000,仅包含字符’a'-’z'。
X.  KMP
http://arc9.riaos.com/?p=5077
http://blog.csdn.net/a1dark/article/details/16994887
咋一看貌似KMP扩展、其实只需要普通的KMP就可以解决、不过需要重新构造一个新字符串C=B+A、于是求出字符串C的next值、最后的值便是answer、
abcde cdefg
==> s3: cdefgabcde, use MKP to compute next or failure function.
int getnext(){
    int j=0,k=-1;
    next[0]=-1;
    int len=strlen(str3);
    while(j<len){
        if(k==-1||str3[j]==str3[k]){
            j++;
            k++;
            next[j]=k;
        }
        else k=next[k];
    }
    int ans=0;
    ans=min(len1,len2);
    ans=min(ans,k);
    return ans;
}
        len1=strlen(str1);
        len2=strlen(str2);
        memset(str3,'',sizeof(str3));
        for(int i=0;i<len2;i++){
            str3[i]=str2[i];
        }
        for(int i=len2,j=0;i<len1+len2,j<len1;i++,j++){
            str3[i]=str1[j];
        }
X. KMP Extended http://blog.csdn.net/xiaozhuaixifu/article/details/12348123
扩展KMP,用extend[i]保存 主串 S[i.....n-1]与 模式串 T的最长公共前缀的长度,其中n是S的长度。
然后扫描一遍 extend[] 如 extend[i] == n-i 那么这个后缀的长度就是我们要求的值。
关于扩展KMP,可以去看论文:《求最长回文子串与最长重复子串》何林 的集训队论文
  1. void getnext(char *T)    
  2. {    
  3.     int k = 0;    
  4.     int Tlen = strlen(T);   
  5.     next[0] = Tlen;    
  6.     while(k < Tlen-1 && T[k] == T[k+1])   
  7.         k++;   
  8.     next[1] = k;    
  9.       
  10.     k = 1;  
  11.     for(int i = 2; i < Tlen; i++)    
  12.     {    
  13.         int p = k+next[k]-1, L = next[i-k];   // p :=已经匹配到的最远的位置  
  14.                                             // L :=     
  15.         if((i-1)+L >= p)    
  16.         {    
  17.             int j=(p-i+1)>0? p-i+1:0;    
  18.             while(i+j < Tlen && T[k+j]==T[j])   
  19.                 j++;    
  20.             next[i] = j;    
  21.             k = i;    
  22.         }    
  23.         else next[k]=L;    
  24.     }    
  25. }    
  26.     
  27. void getextend(char *S,char *T)    
  28. {    
  29.     int a = 0;    
  30.     getnext(T);    
  31.     int Slen = strlen(S);    
  32.     int Tlen = strlen(T);    
  33.     int len = Slen<Tlen ? Slen:Tlen;    
  34.     while(a < len && S[a] == T[a]) a++;    
  35.     extend[0] = a;    
  36.     a = 0;    
  37.     for(int i = 1; i < Slen; i++)    
  38.     {    
  39.         int p = a + extend[a]-1, L = next[i-a];    
  40.           
  41.         if(i+L-1 >= p)  
  42.         {    
  43.             int j = (p-i+1)>0? p-i+1:0;    
  44.               
  45.             while(i+j < Slen && j<Tlen && S[i+j] == T[j])   
  46.                 j++;    
  47.             extend[i] = j;    
  48.             a = i;    
  49.         }    
  50.         else extend[i] = L;    
  51.     }    
  52. }  
  1.         getextend(S,T);  
  2.         int n = strlen(S);  
  3.         int res = 0;  
  4.         for(int i = 0; i < n; i++)  
  5.         {  
  6.             if(extend[i] == n-i)  
  7.             {  
  8.                 res = extend[i];  
  9.                 break;  
  10.             }  
  11.         }  
  12.         cout<<res<<endl; 
X. Brute Force
int judge(char *s,char *t,int low,int high){
05    int index_t = 0;
06    int len_t = strlen(t);
07    while(low <= high)
08    {
09        if(s[low] != t[index_t]) return 0;
10        ++low;
11        ++index_t;
12    }
13    return 1;
14}
15 
16int main()
17{
18    char s[M],t[M];
19    int len_s,len_t,i,flag;
20    while(~scanf("%s%s",s,t))
21    {
22        flag = 0;
23        len_s = strlen(s);
24        len_t = strlen(t);
25        i = len_s > len_t ? len_s-len_t : 0;//减少判断,s串短,从0开始判断,s串长,从len_s-len_t位置开始判断。
26        for(; i < len_s ;++i)
27        {
28            if(s[i] == t[0] && judge(s,t,i,len_s -1))
29            {
30                printf("%d\n",len_s - i);
31                flag = 1;
32                break;
33            }
34        }
35        if(!flag) printf("0\n");
36    }
37    return 0;
38}

X. Rolling Hash
http://blog.csdn.net/xiaozhuaixifu/article/details/12348123
下面介绍一种滚动哈希算法:假设要求S的后缀和T的前缀相等的最大长度,也可以利用滚动哈希在O(n+m)的时间内求解:
假设字符串C = c1c2....cm 定义哈希函数:
H(C) = (c1*b^(m-1)+c2*b^(m-2)+.....+cm*b^(0)) mod h ,
其中 b和h是两个互素的数,这样,我们就可以将字符串 S=s1s2...sn 从位置 k+1 开始长度为m的 子串 S[k+1....k+m]的哈希值
根据 子串 S[k....k+m-1]的哈希值在常数
时间内求出来:
H(S[k+1....k+m]) = H(S[k..k+m-1]*b - sk*b^m +s(k+m)) mod h ,
只要不断这样右移,就可以在O(n)的时间内求得所有位置的哈希值
  1. int overlap(string &a,string &b)  
  2. {  
  3.     int alen = a.size(), blen = b.size();  
  4.     int res = 0;  
  5.     ull ahash = 0, bhash = 0, t = 1;  
  6.     for(int i = 1; i <= min(alen,blen); i++)  
  7.     {  
  8.         ahash = ahash + (a[alen-i]-'a')*t; //a的长度为i的后缀哈希值  
  9.         bhash = bhash*B + b[i-1]-'a';     //b的长度为i的前缀哈希值  
  10.         if(ahash == bhash) res = i;  
  11.         t *= B;   
  12.     }  
  13.     return res;  
http://blog.csdn.net/taoqick/article/details/38470057
  1. int maxLen(const char* S, const char* T) {  
  2.   int Slen = strlen(S), Tlen = strlen(T), i, j, minLen = min(Slen, Tlen), res = 0;  
  3.   typedef unsigned long long ull;  
  4.   ull Shash = 0, Thash = 0, t = 1, radix = 100000007;  
  5.     
  6.   for (int i = 1; i <= minLen; ++i) {  
  7.     Shash = Shash + (S[Slen-i] - 'a') * t;  
  8.     Thash = Thash * radix + (T[i-1] - 'a');      
  9.     t *= radix;  
  10.     if (Shash == Thash)  
  11.       res = i;  
  12.   }  
  13.   return res;  
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