poj1226

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题目链接：http://poj.org/problem?id=1226

题意：就是求n个字符串的最长公共子串，子串是可以反转的

思路：

先对字符串排个序，最短的在最前面。然后通过枚举（从长到短）每一个最短串的子串，看其或其反串是否是其他所有串的子串。

#include <stdio.h>#include <string>#include <iostream>#include <cstdlib>#include <algorithm>#define N 101using namespace std;int P[N],P1[N];bool Comp(string s1, string s2){     return s1.length() < s2.length();}void get_nextval(string ptrn, int plen, int *nextval){     int i = 0;     int j = -1;     nextval[0] = -1;     while(i < plen-1)     {             if(j == -1 || ptrn[i] == ptrn[j])             {                      ++i;                      ++j;                      if( ptrn[i] != ptrn[j] )                          nextval[i] = j;                      else                          nextval[i] = nextval[j];                                       }             else                 j = nextval[j];     } }  int kmp_search(string  src, int slen, string pat, int plen, int const* nextval, int pos) {     int i = pos;     int j = 0;     while( i < slen && j < plen)     {            if( j == -1 || src[i] == pat[j] )            {                ++i;                ++j;                }                   else            {                j = nextval[j];                }     }         if( j >= plen )         return i-plen;     else         return -1; }  int main(int argc, char** argv) {     int t,m,n;     string s[101];     cin>>t;     while(t--)     {         int h;         cin>>h;         for(int i=0; i<h; i++)             cin>>s[i];         sort(s, s+h, Comp);         string a,b;         int len = s[0].length();         int flag = 0;                  for(int i = len; i>0; i--)         {                 for(int j=0; j<=len-i; j++)                 {                         a = s[0].substr(j, i);//                         cout<<"a=" <<a<<endl;                         b = a;                         reverse(b.begin(), b.end());                         m = a.length();                         int k;                         for(k=1; k<h; k++)                         {                              n = s[k].length();//                              cout<<"s[k]="<<k<<"   "<<s[k]<<endl;                              get_nextval(a, m, P);                              get_nextval(b, m, P1);//                              int kmp1 = kmp_search(s[k], n, a, m, P, 0);//                              cout<<kmp1<<endl;//                              int kmp2 = kmp_search(s[k], n, b, m, P1, 0);//                              cout<<kmp2<<endl;                              if( (kmp_search(s[k], n, a, m, P, 0) == -1) && (kmp_search(s[k], n, b, m, P1, 0) == -1) )                                  break;                                }                                                      if(k == h)                         {                              flag = 1;                              break;                              }                     }                 if(flag)break;         }                        if(flag)                 cout<<a.length()<<endl;         else                 cout<<0<<endl;     }         return 0; }

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