hdu1238 Substrings (暴力)

Substrings
Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 10   Accepted Submission(s) : 6
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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

直接枚舉可能的字串，然後進行驗證。

code:

#include <cstdio>  
#include <cstring>  
#include <string>  
#include <iostream>  
using namespace std; 
 
string s[102]; 
int n; 
bool check(string sub) { 
    int i, j; 
    string tmp; 
    for(i=0; i<sub.size(); i++) 
        tmp +=sub[ sub.size()-1-i]; 
    for(i=0; i<n; i++) 
        if(s[i].find(sub)==s[i].npos && s[i].find(tmp)==s[i].npos) 
            return false; 
    return true; 
} 
void solve(int t,int p) { 
    string sub; 
    int i, j, k; 
    int ans =0; 
    for(i=0; i<t; i++) 
        for(j=t-1; j>=i; j--) { 
            if(j-i+1<ans) continue; 
            sub = s[p].substr(i,j-i+1); 
            if(check(sub)) { 
                if(j-i+1>ans) ans = j-i+1; 
            } 
        } 
    cout<<ans<<endl; 
} 
int main() { 
   // freopen("in.txt","r",stdin);  
    int T,i,t,sub_i, j; 
    cin>>T; 
    while(T--) { 
        cin>>n; 
        t = 200; 
        for(i=0; i<n; i++) { 
            cin>>s[i]; 
            if(s[i].size()<t) { 
                t =s[i].size(); 
                sub_i = i; 
            } 
        } 
        solve(t, sub_i); 
    } 
    return 0; 
}