HDOJ 1238 Substrings 【最长公共子串】

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;
    }
}