hdu 1238 Substrings （暴搜，枚举）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1238

Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8391 Accepted Submission(s): 3862



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
There should be one line per test case containing the length of the largest string found.

[align=left]Sample Input[/align]

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

[align=left]Sample Output[/align]

2
2

【题意】

在给定字符串中找到最长公共子串， 或者给定字符串子串的反串；

最多100字符串，最长100，直接枚举

求反串可以用 <algorithm>中的reverse函数 ；

【源代码】

#include <iostream>
#include <algorithm>
using namespace std;
string str[110];
int t;
int cmp(const string& a, const string& b){
return a.length()<b.length();
}
int dfs(string now,string rev,int len){
for(int i=1;i<t;i++){
int sign=0;
for(int j=0;j+len<=str[i].length();j++){
string sub=str[i].substr(j,len); //返回从j 开始 长度为len的子串
if(sub==now||sub==rev)
{
sign=1;break; // 找到就继续找
}
}
if(!sign)
return 0; //否则直接退出
}
return len;
}
int main(){
int n;
cin>>n;

while(n--){
cin>>t;
for(int i=0;i<t;i++)
cin>>str[i];
sort(str,str+t,cmp);
int len=str[0].length();
int max=0;
for(int i=len;i>=1;i--){
for(int j=0;j+i<=len;j++){
string now=str[0].substr(j,i);
string rev=now;
reverse(rev.begin(),rev.end());
//for(int k=now.length()-1;k>=0;k--) //原先手写的翻转字符串，用了上一行的reverse 速度从 140ms 下降到109ms
//	rev+=now[k];
//cout<<rev<<endl;
int ma=dfs(now,rev,i);
if(ma>max)
max=ma;
}
}
cout<<max<<endl;
}
return 0;
}