Substrings HDU - 1238 (kmp+暴力枚舉)

Substrings

HDU - 1238

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.

Input

The 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.

Output

There 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      

暴力枚舉子串，然後用kmp看是否含有這個子串

這裡要求如果逆轉的串是子串也行，是以用到了algorithm檔案中的reverse函數

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
int Next[105];
void getNext(string w,int len){
    int i = -1,j = 0;
    memset(Next,0,sizeof(Next));
    Next[0] = -1;
    while(j < len){
        if(i == -1 || w[i] == w[j]){
            i++,j++;
            Next[j] = i;
        }
        else
            i = Next[i];
    }
}
bool kmp(string w,int m,string s,int n){
    int i = 0,j = 0;
    getNext(w,m);
    while(j < n){
        if(i == -1 || w[i] == s[j])
            i++,j++;
        else
            i = Next[i];
        if(i >= m){
            return true;
        }
    }
    return false;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        string str[102];
        int ans = 0;
        int n,i,j,k;
        scanf("%d",&n);
        for(i = 0; i < n; i++){
            cin >> str[i];
        }
        for(i = 0; i < str[0].length(); i++){
            for(j = i; j < str[0].length(); j++){//j從i開始，長度為1的串也要算
                string w = str[0].substr(i,j-i+1);
                for(k = 1; k < n; k++){
                    string rw = w;
                    reverse(w.begin(),w.end());
                    if(!kmp(w,w.length(),str[k],str[k].length())&&!kmp(rw,rw.length(),str[k],str[k].length()))
                        break;
                }
                if(k >= n){
                    if(w.length()>ans)
                        ans = w.length();
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}