HDU 1238 Substrings（暴力+stl）

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

23ABCDBCDFFBRCD2roseorchid

Sample Output

22

题解：

给你n个串，让你找这些串的最长公共子串

直接找出最短的串，枚举他的所有子串，然后暴力遍历，stl很方便，就是也有一些坑，莫名其妙不知道错哪

不过stl用起来很慢，不然c风格的字符串。。

虽然我没用到一个函数，但是还是学习一下

包含文件：string.h

 函数名: strstr
函数原型：extern char *strstr(char *str1, char *str2);
功能：找出str2字符串在str1字符串中第一次出现的位置（不包括str2的串结束符）。

如果没找到就会返回NULL

代码：

#include<algorithm>#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<vector>#include<deque>using namespace std;#define lson k*2#define rson k*2+1#define M (t[k].l+t[k].r)/2#define INF 1008611111#define ll long long#define eps 1e-15int vis[20005];int cmp1(string x,string y)//从小到大{    return x.size()<y.size();}int cmp2(string x,string y)//从大到小{    return x.size()>y.size();}int main(){    int test,n,i,j,k,cent,d,len,tot;    scanf("%d",&test);    while(test--)    {        string p[105],ss[20005];        scanf("%d",&n);        cent=0;        for(i=0;i<n;i++)        {            cin>>p[i];            vis[i]=0;        }        sort(p,p+n,cmp1);        for(i=0;i<p[0].size();i++)//枚举子串        {            for(j=1;i+j<p[0].size()+1;j++)            {                ss[cent]=p[0].substr(i,j);                cent++;                tot=0;                for(k=ss[cent-1].size()-1;k>=0;k--,tot++)                {                    ss[cent].push_back(ss[cent-1][k]);                }                cent++;            }        }        sort(ss,ss+cent,cmp2);        cent=unique(ss,ss+cent)-ss;//可以去一下重        len=0;        for(i=0;i<cent;i++)        {            int tag=1;            for(j=1;j<n;j++)            {                if(p[j].find(ss[i])==string::npos)//如果没找到                {                    tag=0;                    break;                }            }            if(tag)            {                len=ss[i].size();                break;            }        }        printf("%d\n",len);    }    return 0;}