hdu 1238 字符串处理 Substrings

Substrings

TimeLimit: 2000/1000 MS (Java/Others)    Memory Limit:65536/32768 K (Java/Others)
Total Submission(s): 9134    Accepted Submission(s): 4312

Problem Description

You are given a number ofcase-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 thegiven strings.

 

 

Input

The first line of the input filecontains 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 casecontains a single integer n (1 <= n <= 100), the number of given strings,followed by n lines, each representing one string of minimum length 1 andmaximum length 100. There is no extra white space before and after a string.

 

 

Output

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

 

 

Sample Input

2

3

ABCD

BCDFF

BRCD

2

rose

orchid

 

 

Sample Output

2

2

 

 

 

题意：找出一个子串x  使x或者x的逆串使者n个字符串的公共子串 输出x的最大的长度

分析：

要想找到公共子串那么只需找到最短的字符串  然后枚举所有的子串 然后在一个一个找   还好范围是100 不会超的 

AC代码：

#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);   ///t组测试数据

        while(t--) 

        { 

            scanf("%d",&n);  ///n个字符串

            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]);   ///最小串的长度

            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];    ///s1正序  s2倒序

                        s2[j-k] =str[f][k]; 

                    } 

                    s1[j-i+1] = s2[j-i+1] ='\0';  ///'\0'之前是s1串和s2串的实际长度 

                    int l = strlen(s1); 

                    for(k = 0;k<n;k++)///枚举所有串 

                    { 

                        ///strstr(s,s1)  在s串中找s1串出现的位置 返回值 bool型 找到 true 否则 fal

                        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; 

    }