HDOJ 1238 Substrings

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9011    Accepted Submission(s): 4245

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

 

题意：给我们n个字符串，问我们其中三个的最长公共子序列或者反序列长度是多少，由于这里的数据很弱，我们可以直接借用C++ String类型的一些库函数进行暴力求解~。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
int n;
const int maxn = 100+10;
string str[maxn];
int cmp(string a,string b)
{
	return a.length() < b.length();
}
int dfs(string &s)
{
	int len = s.length();
	int flag = 1;
	for(int i=len; i>=1; i--)
		for(int j=0; i+j<=len; j++)
		{
			string temp = s.substr(j,i);
			string t = temp;
			string retemp = t.assign(t.rbegin(),t.rend());
			for(int k=1; k<n; k++)
				if(str[k].find(temp) == -1 && str[k].find(retemp) == -1)
				{
					flag = 0;
					break;
				}
			if(flag) return temp.length();
			flag = 1;
		}
	return 0;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(int i=0; i<n; i++)
			cin >> str[i];
		sort(str,str+n,cmp);
		int ans = dfs(str[0]);
		printf("%d\n",ans);
	}
}