hdu 1238 Substrings

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7261    Accepted Submission(s): 3277

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

算法解析：

在多个字符串中寻找最长字串，并输出该字串的长度。首先找出最短的一个字符串，以最短字符串的所有子串从长到短搜索所有字符串，若全部满足存在某个最短字符串的子串或反转子串，输出该子串的长度。

需要利用的字符串函数

strlen：计算字符串的长度
strncpy：复制字符串的子串
strcpy：复制字符串
strstr：在字符串中寻找子字符串
strrev：对字符串进行反序

代码如下：

#include<stdio.h>#include<string.h>char str[100][101];int n;int searchMaxSubString(char *source){int i, j, subStrLen,sourceStrLen;int foundMaxSubStr;char subStr[101], revSubStr[101];subStrLen= sourceStrLen=strlen(source) ;while ( subStrLen > 0 ){                                                                    //搜索不同长度的子串，从最长的子串开始搜索for (i = 0; i <= sourceStrLen - subStrLen; i++) {                                                                //搜索长度为subStrLen 的全部子串，共 sourceStrLen – subStrLen个                                                             //取source字符串中第i个字符开始的长度为subStrLend的子串strncpy(subStr, source+i, subStrLen);strncpy(revSubStr, source+i, subStrLen);subStr[subStrLen] = revSubStr[subStrLen] = '\0';strrev(revSubStr);                                           //得到反序子串foundMaxSubStr = 1;for( j = 0; j < n; j++)if ( strstr(str[j], subStr) == NULL && strstr(str[j], revSubStr) == NULL ) {foundMaxSubStr = 0;break;}if (foundMaxSubStr)return(subStrLen);}subStrLen--;}return 0;}int main(){int N,min,a,x,i;scanf("%d",&N);while(N--){scanf("%d",&n);min=100,i=0;while(i<n){scanf("%s",str[i]);a=strlen(str[i]);if(min>a){min=a;x=i;}i++;}printf("%d\n",searchMaxSubString(str[x]));}return 0;}