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最新推荐文章于 2020-09-26 14:57:02 发布

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最新推荐文章于 2020-09-26 14:57:02 发布

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分类专栏： ACM算法_KMP 文章标签： KMP算法

本文链接：https://blog.youkuaiyun.com/zhang__liuchen/article/details/78558857

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本文介绍了一个关于寻找多个字符串中最长公共子串或其反转子串的问题，并提供了一段C++代码实现。该算法首先确定最短字符串，然后通过暴力匹配所有可能的子串来找出最长的有效子串。

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Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11034 Accepted Submission(s): 5310

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

找到最短的串，暴力所有字串，去匹配。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
int main()
{
    int ca,n,f;
    scanf("%d",&ca);
    string s[105];
    while(ca--)
    {
        int mma=1000;
        int ma=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            cin>>s[i];
            if(s[i].size()<mma)
            {
                mma=s[i].size();
                f=i;
            }
        }
        int le=s[f].size();
        for(int i=0;i<le;i++)
        {
            for(int j=0;j<le;j++)
            {
                string s1,s2;
                s1=s[f].substr(i,j);
                s2=s1;
                reverse(s2.begin(),s2.end());
                int flag=1;
                for(int k=0;k<n;k++)
                {
                    if(s[k].find(s1,0)==-1&&s[k].find(s2,0)==-1)
                    {
                        flag=0;
                        break;
                    }
                }
                if(flag&&s1.size()>ma)
                    ma=s1.size();
            }
        }
        cout<<ma<<endl;
    }
    return 0;
}