HDU - 2018杭电ACM集训队单人排位赛 - 2 - Problem D. Team Name

Problem D Team Name

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

Total Submission(s): 129    Accepted Submission(s): 68

Problem Description

After all the teams have been matched, what to do next is of course to think about a nice team name. Now it is known that there are n teams taking part in the Guangxi Province Collegiate Programming Contest. And the name of every team is a string consists of English lower characters. Now Luras needs to get her team name, she doesn’t want the name be any consecutive substring of any other teams. And she prefers shorter names. If there are many choices with the shortest length, she would prefer the one which is the smallest lexicographically. Now could you decide the team name for her? We regard string a is lexicographically smaller than string b if there exists such index j that a[i] == b[i] for all i < j and a[j] < b[j].

Input

The first line is an integer T which indicates the case number.

And as for each case,  there will be n + 1 lines.

In the first line, there is one integer n, which indicates the number of teams.

Then there will be n strings of the next n lines, indicate the name of every team in each line.

It is guaranteed that——

T is about 100.

for 100% cases, 1 <= n <= 100, 1 <= |s|（the length of s）<= 30.

Output

As for each case, you need to output a single line.

There should be one string in the line which means the name Luras will give to her team.

Sample Input

2

3

a

b

c

2

abcdefghijklmnopqrstuvwxyz

aa

Sample Output

d

ac

解题思路：不是 n 个队名中的子串 && 队名字典序尽可能小。

AC 代码

/

#include<bits/stdc++.h>
#include<cmath>
#define mem(a,b) memset(a,b,sizeof a);
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
char a[110][110], ans[110];
bool ok;
int n;
void dfs(int dep,int len)
{
    if(len==dep)
    {
        ans[len++]='\0';
        for(int i=0;i<n;i++)
            if(strstr(a[i],ans)) return;
        ok=true;
        return;
    }
    for(int i=0;i<26;i++)
    {
        ans[dep]=i+'a';
        dfs(dep+1,len);
        if(ok) return;
    }
    return;
}
int main()
{
    int T; scanf("%d",&T);
    while(T-- && ~scanf("%d",&n))
    {
        for(int i=0;i<n;i++) scanf("%s",a[i]);
        ok=false;
        for(int len=1;;len++)
        {
            if(ok) break;
            dfs(0,len);
        }
        printf("%s\n",ans);
    }
    return 0;
}