HDU1238-Substrings

来源：互联网发布：mac 更新java版本编辑：程序博客网时间：2024/05/19 00:15

Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9866 Accepted Submission(s): 4679

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

Author

Asia 2002, Tehran (Iran), Preliminary

题思：找出所有字符串中共同拥有的一个子串，该子串（正、逆字符）是任何一个母串的子串，求该子串的最长长度。

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <cmath>#include <algorithm>using namespace std;const int INF=0x3f3f3f3f;int main(){    int t,n;    char str[105][105],s1[105],s2[105];    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        int mi=INF,p;        for(int i=0; i<n; i++)        {            scanf("%s",str[i]);            int len=strlen(str[i]);            if(mi>len)            {                mi=len;                p=i;            }        }        int len1=strlen(str[p]);        int flag=1;        int ma=0;        for(int i=0;i<len1;i++)        {            for(int j=i;j<len1;j++)            {                flag=1;                for(int k=i;k<=j;k++)                {                    s1[k-i]=str[p][k];                    s2[j-k]=str[p][k];                }                s1[j-i+1]=s2[j-i+1]='\0';                int len2=strlen(s1);                for(int k=0; k<n; k++)                {                    if(!strstr(str[k],s1)&&!strstr(str[k],s2))                    {                        flag=0;                        break;                    }                }                if(len2>ma&&flag)                    ma=len2;            }        }        printf("%d\n",ma);    }    return 0;}

0 0