kuangbin专题十六 KMP&&扩展KMP HDU1238 Substrings

You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.

InputThe first line of the input file contains a single integer t
(1 <= t <= 10), the number of test cases, followed by the input
data for each test case. The first line of each test case contains a
single integer n (1 <= n <= 100), the number of given strings,
followed by n lines, each representing one string of minimum length 1
and maximum length 100. There is no extra white space before and after a
string.

OutputThere should be one line per test case containing the length of the largest string found.

Sample Input

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

Sample Output

2
2
 
这道题和之前一道题类似，都是多个字符串匹配问题，只是这道题多了倒序的匹配。直接reverse就可以了
遍历第一个串的所有后缀，然后匹配每个字符串的正反序。得到公共匹配的最小。然后枚举所有后缀取最大

 #include<stdio.h>
 #include<iostream>
 #include<string>
 #include<algorithm>
 #include<vector>
 using namespace std;
 int _,n,Next[];
 vector<string> t;
 
 void prekmp(string s) {
     int len=s.size();
     int i,j;
     j=Next[]=-;
     i=;
     while(i<len) {
         while(j!=-&&s[i]!=s[j]) j=Next[j];
         Next[++i]=++j;
     }
 }
 
 int kmp(string t,string p) {
     int lent=t.size(),lenp=p.size();
     int i=,j=,ans=-,res;
     res=-;
     while(i<lent) {
         while(j!=-&&t[i]!=p[j]) j=Next[j];
         ++i;++j;
         res=max(res,j);
     }
     ans=max(ans,res);
     reverse(t.begin(),t.end());
     res=-;
     i=,j=;
     while(i<lent) {
         while(j!=-&&t[i]!=p[j]) j=Next[j];
         ++i;++j;
         res=max(res,j);
     }
     ans=max(ans,res);
     return ans;
 }
 
 int main() {
    // freopen("in","r",stdin);
     for(scanf("%d",&_);_;_--) {
         scanf("%d",&n);
         string s;
         for(int i=;i<n;i++) {
             cin>>s;
             t.push_back(s);
         }
         string str=t[],tempstr;
         int len=str.size(),maxx=-;
         for(int i=;i<len;i++) {
             tempstr=str.substr(i,len-i);
             int ans=0x3f3f3f;
             prekmp(tempstr);
             for(int j=;j<t.size();j++) {
                 ans=min(kmp(t[j],tempstr),ans);
             }
             maxx=max(maxx,ans);
         }
         printf("%d\n",maxx);
         t.clear();
     }
 }