HDOJ 1238：Substrings 寻找最长子序列 解题报告

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4177    Accepted Submission(s): 1871

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
   
   
   
   

 

题目简意：寻找给定若干字符串中的最长公共子序列 其中该序列可正序可反序

// HDOJ_1238.cpp : Substrings
//求子串：s.substr(position,len);
//查找字串：s.find(s1);如果没找到，则返回-1，可用s.find(s1)!=-1来判断是否找到子串
//求字符串长度：s.length()
//翻转字符串：reverse(s.begin(),s.end())会把s自己本身翻转过来
#include "StdAfx.h"
#include <fstream>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>	
//reverse函数在algorithm中 不过VS有了"StdAfx.h"就不需要它了
//sort也在algorithm中 但只有"StdAfx.h"又不行 奇怪
using namespace std;
int main()
{
	ifstream cin("1238_input.txt");	//调试用
	int t, n, i, j, k;	//t=test cases
	cin >> t;
	while ( t-- ){
		cin >> n;
		string *a = new string[n];
		string *b = new string[n];
		int minLength = 150;
		int minIndex;
		for( i=0; i<n; i++ ){
			cin >> a[i];
			b[i] = a[i];
			reverse(b[i].begin(),b[i].end());	//reverse为void类型
			if( minLength > a[i].length() ){
				minLength = a[i].length();
				minIndex = i;
			}
		}
		//cout << minLength << endl;
		string subString;
		bool fail;
		int maxLength=0;
		for( i=0; i<minLength; i++ )
			for( j=1; j<=minLength-i; j++ ){
				fail = 0;
				subString = a[minIndex].substr(i,j);
				//cout << subString << endl;
				for( k=0; k<n; k++ ){
					if(a[k].find(subString)!=-1 && k==n-1 ){
						maxLength = max(maxLength, j);
						//cout << maxLength << endl;
					}
					if(a[k].find(subString)==-1){	//返回-1表未找到
						if(b[k].find(subString)==-1 ){
							fail = 1;
							break;	//从k循环跳出
						}
						else if( k==n-1 )
							maxLength = max(maxLength, j);
					}
				}
				if( fail == 1 )
					break;
			}

		delete []a;
		delete []b;
		cout << maxLength << endl;
	}//while
    return 0;
}