POJ 1226 Substrings

Substrings

Time Limit: 1000ms

Memory Limit: 10000KB

This problem will be judged on PKU. Original ID: 1226
64-bit integer IO format: %lld Java class name: Main

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 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

Source

Tehran 2002 Preliminary

解题：霸蛮好了。。。

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <cmath>

 #include <algorithm>

 #include <climits>

 #include <vector>

 #include <queue>

 #include <cstdlib>

 #include <string>

 #include <set>

 #include <stack>

 #define LL long long

 #define pii pair<int,int>

 #define INF 0x3f3f3f3f

 using namespace std;

 string str[];

 int main() {

     int t,i,j,k,n;

     bool  flag;

     scanf("%d",&t);

     while(t--){

         scanf("%d",&n);

         for(i = ; i < n; i++)

             cin>>str[i];

         sort(str,str+n);

         flag = false;

         for(k = str[].length(); k; k--){

             for(i = ; i + k <= str[].length(); i++){

                 string a = str[].substr(i,k);

                 string b(a.rbegin(),a.rend());

                 for(j = ; j < n; j++)

                     if(str[j].find(a) == - && str[j].find(b) == -) break;

                 if(j == n) {flag = true;break;}

             }

             if(flag) break;

         }

         flag?printf("%d\n",k):puts("");

     }

     return ;

 }

巴特西