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最新推荐文章于 2019-07-19 11:11:50 发布

NineFailure

最新推荐文章于 2019-07-19 11:11:50 发布

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分类专栏： ACM练习字符串hash

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天坑哥希望为他的餐厅取一个不落俗套的英文名字，并且要确保这个名字与现有名称的所有子串都不相同。面对仅有八种字符的选择限制，如何找到最短且字典序最小的独特名称？本文通过算法解决这一问题。

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After improving the marketing strategy, TIANKENG has made a fortune and he is going to step into the status of TuHao. Nevertheless, TIANKENG wants his restaurant to go international, so he decides to name his restaurant in English. For the lack of English skills, TIANKENG turns to CC, an English expert, to help him think of a property name. CC is a algorithm lover other than English, so he gives a long string S to TIANKENG. The string S only contains eight kinds of letters——-‘A’, ‘B’, ‘C’, ‘D’, ‘E’, ‘F’, ‘G’, ‘H’. TIANKENG wants his restaurant’s name to be out of ordinary, so the restaurant’s name is a string T which should satisfy the following conditions: The string T should be as short as possible, if there are more than one strings which have the same shortest length, you should choose the string which has the minimum lexicographic order. Could you help TIANKENG get the name as soon as possible?

Meanwhile, T is different from all the substrings of S. Could you help TIANKENG get the name as soon as possible?
Input
The first line input file contains an integer T(T<=50) indicating the number of case.
In each test case:
Input a string S. the length of S is not large than 1000000.
Output
For each test case:
Output the string t satisfying the condition.(T also only contains eight kinds of letters——-‘A’, ‘B’, ‘C’, ‘D’, ‘E’, ‘F’, ‘G’, ‘H’.)
Sample Input
3
ABCDEFGH
AAABAACADAEAFAGAH
ACAC
Sample Output
AA
BB
B

首先只要确定他这个一百万的长度到几就不能枚举所有情况就好
是6

然后开始枚举。。

#include <iostream>
#include <cstring>
#include<algorithm>
#include <cstdio>
#include<vector>
#include<queue>
#include<string>
#include<map>
#include<stack>
using namespace std;
long long ww[113];
long long T;
bool mp[10000000];
int main()
{
//#define int long long 
    int T;
    cin >> T;
    ww[0] = 1;
    for (int a = 1; a < 14; a++)ww[a] = ww[a - 1] * 9;
    string q;
    while (T--)
    {
        memset(mp, 0, sizeof(mp));
        int ss = 1;
        cin >> q;
        int we = min(6, (int)q.size());
        int jjj = 0;
        for (int a = 1; a <= we; a++)
        {
            for (int b = 0; b < q.size(); b++)
            {
                if (b + a - 1 >= q.size())break;
                int sd = 0;
                for (int c = b; c <= b + a - 1; c++)
                {
                    sd *= 9;
                    sd += q[c] - 'A'+1;
                }
                mp[sd] = 1;
                if (jjj)break;
            }
            for (int d = ww[a-1];d<=ww[a]; d++)
            {
                int rt = d;
                int jcjc = 0;
                while (rt)
                {
                    if (rt % 9 == 0)
                    {
                        jcjc = 1;
                        break;
                    }
                    rt /= 9;
                }
                if (jcjc)continue;
                if (mp[d])continue;
                ss = d;
                jjj = 1;
                break;
            }
            if (jjj)break;
        }
        stack<int>sd;
        while (ss)
        {
            sd.push(ss % 9);
            ss /= 9;
        }
        while (!sd.empty())
        {
            printf("%c", (char)sd.top() + 'A'-1);
            sd.pop();
        }
        printf("\n");
    }
}