HDOJ1238 C++字符串函数水题

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10797    Accepted Submission(s): 5167

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

 

Author

Asia 2002, Tehran (Iran), Preliminary

 

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#include <ctime>#include <iostream>#include <cmath>#include <cstring>#include <algorithm>using namespace std;const int maxn = 105;int n,i,j,k;string str[maxn];bool cmp(string a, string b) {  return a.length()<b.length();}void init(){   cin >> n;   for (i=1; i<=n; i++) cin >> str[i];   sort(str+1,str+n+1,cmp);}int Find(){    int tmp,len;    bool flag;    string s = str[1];    len = s.length();    string temp,retmp;    for (i=len; i>=1; i--) {        for (j=0;j+i<=len; j++) {            flag = 1;            temp = s.substr(j,i); //从第j位开始复制长度i的字符串            retmp = temp;            reverse(retmp.begin(),retmp.end()); //反转            for (k=2; k<=n; k++)                if (str[k].find(temp)==-1 && str[k].find(retmp)==-1) {                  flag = 0;                  break;                }            if (flag) return i;        }    }    return 0;}int main(){    std::ios::sync_with_stdio(false);    int t;    cin >> t;    while (t--) {        init();        cout << Find() << endl;    }    return 0;}