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POJ 1226 Substrings

时间:2014-08-23 21:34:21      阅读:236      评论:0      收藏:0      [点我收藏+]

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Substrings

Time Limit: 1000ms
Memory Limit: 10000KB
This problem will be judged on PKU. Original ID: 1226
64-bit integer IO format: %lld      Java class name: Main
 
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 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 

Source

 
 
解题:霸蛮好了。。。
 
bubuko.com,布布扣
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 string str[110];
18 int main() {
19     int t,i,j,k,n;
20     bool  flag;
21     scanf("%d",&t);
22     while(t--){
23         scanf("%d",&n);
24         for(i = 0; i < n; i++)
25             cin>>str[i];
26         sort(str,str+n);
27         flag = false;
28         for(k = str[0].length(); k; k--){
29             for(i = 0; i + k <= str[0].length(); i++){
30                 string a = str[0].substr(i,k);
31                 string b(a.rbegin(),a.rend());
32                 for(j = 1; j < n; j++)
33                     if(str[j].find(a) == -1 && str[j].find(b) == -1) break;
34                 if(j == n) {flag = true;break;}
35             }
36             if(flag) break;
37         }
38         flag?printf("%d\n",k):puts("0");
39     }
40     return 0;
41 }
View Code

 

POJ 1226 Substrings

标签:style   blog   http   color   java   os   io   for   ar   

原文地址:http://www.cnblogs.com/crackpotisback/p/3931725.html

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