hdu1238 Substrings

最新推荐文章于 2019-07-14 21:53:29 发布

原创最新推荐文章于 2019-07-14 21:53:29 发布 · 451 阅读

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本文介绍了一种算法，用于从一组给定的字符串中找到最长的字符串X，该字符串或者其逆序能作为子串出现在任意给定的字符串中。通过排序和深度优先搜索等方法实现。

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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

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

Sample Output

2
2

思路：这题如果自己写字符串的倒序，挺麻烦的。我们要善于运用c++提供的api,string里面有很多有用的东西，别忘了。
给字符串排序，自己写个compare函数。从小到大排序，取出最小的字符串，遍历所有它的字串，再求出倒序字串。最后在各字符串看能否找到。

代码：
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
string ss[200];
int com(const string p, const string q)
{
    return p.length() < q.length();
}

int dfs(string & str, int n)
{
    int len,i,j,k,maxx;
    len = str.length();
    int flag = 1;
    //int ant = 0;
    for (i = len; i >= 1; i--)//长度
    {
        for(j = 0; j+i <= len; j++)//起点
        {
            string sub = str.substr(j,i);
            string ww = sub;//必须有个中间变量
            string rsub = ww.assign(ww.rbegin(),ww.rend());//使用rbegin,rend时ww已经改变
            for(k = 1; k < n; k++)
            {
                if (ss[k].find(sub) == -1 && ss[k].find(rsub) == -1)
                {
                    flag = 0;
                    break;
                }
            }
            if (flag)
            {
                maxx = sub.length();
                return maxx;
            }
            flag = 1;
        }
    }
    return 0;

}
int main()
{
   int t,n,i;
   scanf("%d",&t);
   while (t--)
   {
       scanf("%d",&n);
       for (i = 0; i < n; i++)
       {
           cin >> ss[i];
       }
       sort(ss,ss+n,com);
       cout << dfs(ss[0],n) << endl;

   }
   return 0;
}