HDOJ 1238 Substrings 【最长公共子串】

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

Total Submission(s): 11430 Accepted Submission(s): 5490

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

题意

给出一系列字符串 求出这些字符串中的最长公共子串

思路

可以用C++ STL 里面的东西 去找子串

因为题目要求 是可以逆字符串的

所以可以用reverse

然后 查找可以用find

AC代码

#include <iostream>

#include <cstdio>

#include <algorithm>

#include <cmath>

#include <deque>

#include <vector>

#include <queue>

#include <string>

#include <cstring>

#include <map>

#include <stack>

#include <set>

#include <cstdlib>

#include <ctype.h>

#include <numeric>

#include <sstream>

using namespace std;

typedef long long LL;

const double PI = 3.14159265358979323846264338327;

const double E = 2.718281828459;

const double eps = 1e-6;

const int MAXN = 0x3f3f3f3f;

const int MINN = 0xc0c0c0c0;

const int maxn = 1e2 + 5;

const int MOD = 1e9 + 7;

string s[maxn];

int main()

{

    int t;

    cin >> t;

    while (t--)

    {

        int n;

        cin >> n;

        int i, j, k;

        int sub;

        int len = MAXN;

        for (i = 0; i < n; i++)

        {

            cin >> s[i];

            if (s[i].size() < len)

            {

                len = s[i].size();

                sub = i;

            }

        }

        int ans = 0;

        for (i = s[sub].size(); i > 0; i--)

        {

            for (j = 0; j < s[sub].size() - i + 1; j++)

            {

                string s1, s2;

                s1 = s[sub].substr(j, i);

                s2 = s1;

                reverse(s2.begin(), s2.end());

                for (k = 0; k < n; k++)

                {

                    if (s[k].find(s1, 0) == -1 && s[k].find(s2, 0) == -1)

                        break;

                }

                if (k == n && s1.size() > ans)

                    ans = s1.size();

            }

        }

        cout << ans << endl;

    }

}