Substrings（hdu1238）字串匹配

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Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7205 Accepted Submission(s): 3255

Problem DescriptionYou 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. 

 

InputThe 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.  

 

OutputThere should be one line per test case containing the length of the largest string found. 

 

Sample Input23ABCDBCDFFBRCD2roseorchid 

 

Sample Output22  

這個找字串的問題，題目大概意思就是找出所有字串中共同擁有的一個子串，

該子串（正、逆字元）是任何一個母串的子串，求該子串的最長長度。

想到用STRING裡的成員函數和STL的reverse反轉函數，

思路:

先找出最短的母串，即該符合要求的子串肯定在這個母串中，即在從長到短

從最短母串中取子串，在子串正反去查看是否符合要求。

說實話今天又學到了一些知識，我表示這C++的很多函數真shi 強大啊。

 

 

ps:http://acm.hdu.edu.cn/showproblem.php?pid=1238

 

 

 

#include<iostream>#include<string>#include<algorithm>//STL reverse函數的標頭檔，reverse反轉函數，using namespace std;int main(){    int cas,len,sub,maxn;    int n,k,i,j;    string s[102];    cin>>cas;    while(cas--)    {        cin>>n;        len=1000;        sub=0;        for(i=0; i<n; i++)        {            cin>>s[i];            if(len>s[i].size())//找最小 的母串            {                len=s[i].size();                sub=i;            }        }        maxn=0;        for(i=s[sub].size(); i>0; i--) //從最小的母串開始從長到短找子串，        {            for(j=0; j<s[sub].size()-i+1; j++) //長度為i的子串在母串中找            {                string s1,s2;//s1為子串正 ，s2為子串反                s1=s[sub].substr(j,i);//去j開始i長度是字元                s2=s1;                reverse(s2.begin(),s2.end());//反串                for( k=0; k<n; k++)                {                    if(s[k].find(s1,0)==-1&&s[k].find(s2,0)==-1) //當正反子串在母串中都未發現時即跳出                        break;                }                if(k==n&&maxn<s1.size())                    maxn=s1.size();            }        }        cout<<maxn<<endl;    }    return 0;}