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[110];
int main() {
    int t,i,j,k,n;
    bool  flag;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(i = 0; i < n; i++)
            cin>>str[i];
        sort(str,str+n);
        flag = false;
        for(k = str[0].length(); k; k--){
            for(i = 0; i + k <= str[0].length(); i++){
                string a = str[0].substr(i,k);
                string b(a.rbegin(),a.rend());
                for(j = 1; j < n; j++)
                    if(str[j].find(a) == -1 && str[j].find(b) == -1) break;
                if(j == n) {flag = true;break;}
            }
            if(flag) break;
        }
        flag?printf("%d
",k):puts("0");
    }
    return 0;
}