hdu1238 Substrings (暴力)

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

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

直接枚举可能的字串，然后进行验证。

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;
}