HDU1238：Substrings

Problem Description

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

23ABCDBCDFFBRCD2roseorchid

 

Sample Output

22

 

 

题意：找出所有串的最长的公共连续子串

思路：直接从最小的那串，枚举所有子串去寻找，反正最多100串，最长100字符，大胆的枚举吧！骚年！

 

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int main(){    int t,n,i,j,k,MIN,f,len,MAX;    char str[105][105],s1[105],s2[105];    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        MIN = 1000;        for(i = 0; i<n; i++)        {            scanf("%s",str[i]);            len = strlen(str[i]);            if(MIN>len)//找到最小串            {                MIN = len;                f = i;            }        }        len = strlen(str[f]);        int flag = 1;        MAX = 0;        for(i = 0;i<len;i++)//作为标本串子串的头        {            for(j = i;j<len;j++)//子串的尾            {                for(k = i;k<=j;k++)//复制为两个串，顺序串s1，逆序串s2                {                    s1[k-i] = str[f][k];                    s2[j-k] = str[f][k];                }                s1[j-i+1] = s2[j-i+1] = '\0';                int l = strlen(s1);                for(k = 0;k<n;k++)//枚举所有串                {                    if(!strstr(str[k],s1) && !strstr(str[k],s2))                    {                        flag = 0;                        break;                    }                }                if(l>MAX && flag)                MAX = l;                flag = 1;            }        }        printf("%d\n",MAX);    }    return 0;}