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; }