Substrings

Substrings

时间限制(普通/Java):3000MS/10000MS          运行内存限制:65536KByte
总提交:60            测试通过:32

描述

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.

输入

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.

输出

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

样例输入

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

样例输出

2
2

思路： 先找出最短的母串，即该符合要求的子串肯定在这个母串中，即在从长到短
从最短母串中取子串，在子串正反去查看是否符合要求。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){    string s[200];    int n,m,i,j,min1,minline,k;    scanf("%d",&n);    while(n--)    {        int max1=0;        min1=10000;        scanf("%d",&m);        getchar();        for(i=0; i<m; i++)        {            cin>>s[i];            int len=s[i].size();            if(len<min1)            {                min1=len;                minline=i;            }        }       // printf("%d\n",min1);        for(i=s[minline].size(); i>=0; i--)        {            for(j=0; j<=i; j++)            {                string s1,s2;                s1 = s[minline].substr(j, i);                s2=s1;                reverse(s2.begin(), s2.end());                for(k=0; k<m; k++)                {                    if(s[k].find(s1)==-1&&s[k].find(s2)==-1)                    {                        break;                    }                }                if(k==m&&max1<s1.size())                {                    max1=s1.size();                }            }        }        printf("%d\n",max1);    }    return 0;}