hdu 1238 Substrings(暴力枚举)

最新推荐文章于 2020-03-25 18:03:17 发布

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本文探讨了一个字符串处理问题——寻找能在多个给定字符串中找到的最长子串（包括其反转形式），并提供了一种通过枚举和匹配来解决该问题的方法。

Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11496 Accepted Submission(s): 5521

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

Author

Asia 2002, Tehran (Iran), Preliminary

Recommend

Eddy

题意：

给了N个串，问最长的那个串满足出现在所有的串，这个串既可以是正的也可以是翻转的出现在其他串中。求最长的串的长度。

思路：

因为数据量比较小，可以暴力枚举第一个串中的所有子串然后在其余串中进行匹配，并且记录下能出现在所有串中的那个最大长度。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
string str[105];
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            cin>>str[i];
        }
        int len=str[1].length();
        int ans=0;
        for(int i=0;i<len;i++)
        {
            for(int j=1;j<=len-i;j++)
            {
                string a=str[1].substr(i,j);
                string b=a;
                reverse(b.begin(),b.end());
                int f=1;
                for(int k=1;k<=n;k++)
                {
                    if(str[k].find(a)==-1&&str[k].find(b)==-1)
                    {
                        f=0;
                        break;
                    }
                }
                if(f)
                {
                    ans=max(ans,j);
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}