HDU 3294Life Forms

Posted 2022-10-22 rentu

Description

You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour, wrinkles, ears, eyebrows and the like. A few bear no human resemblance; these typically have geometric or amorphous shapes like cubes, oil slicks or clouds of dust.

The answer is given in the 146th episode of Star Trek - The Next Generation, titled The Chase. It turns out that in the vast majority of the quadrant‘s life forms ended up with a large fragment of common DNA.

Given the DNA sequences of several life forms represented as strings of letters, you are to find the longest substring that is shared by more than half of them.

Input

Standard input contains several test cases. Each test case begins with 1 ≤ n ≤ 100, the number of life forms. n lines follow; each contains a string of lower case letters representing the DNA sequence of a life form. Each DNA sequence contains at least one and not more than 1000 letters. A line containing 0 follows the last test case.

Output

For each test case, output the longest string or strings shared by more than half of the life forms. If there are many, output all of them in alphabetical order. If there is no solution with at least one letter, output "?". Leave an empty line between test cases.

Sample Input

3
abcdefg
bcdefgh
cdefghi
3
xxx
yyy
zzz
0

Sample Output

bcdefg
cdefgh

?

【题目大意】

给定多个字符串，寻找字符串中出现的最长相同子串，如果有多个全部输出

 

【解题思路】

首先求后缀数组，然后二分查找最长长度的子串

然后在二分答案的时候，记录答案

 

【代码细节】

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<algorithm>
  4 #include<cmath>
  5 #include<vector>
  6 #include<cstring>
  7 using namespace std;
  8 const int MAXN = 110;
  9 const int MAXM = 2010;
 10 int n = 0, num;
 11 int down;
 12 int s[MAXN*MAXM];
 13 vector<int> ret;
 14 char tmp[MAXM];
 15 int  SA[MAXN*MAXM], height[MAXN*MAXM];
 16 int  rak[MAXN*MAXM],k;
 17 int  c[MAXN*MAXM], x[MAXN*MAXM], y[MAXN*MAXM];
 18 int  pos[MAXN],id[MAXN*MAXM];
 19 vector<int> ans;
 20 bool vis[MAXN];
 21 void init()
 22 
 23     memset(c, 0, sizeof(c));
 24     memset(x, 0, sizeof(x));
 25     memset(y, 0, sizeof(y));
 26     return;
 27 
 28 void get_SA(int *s)
 29 
 30     int m = down;
 31     for (int i = 1; i <= n; i++)
 32         ++c[x[i] = s[i]];
 33     for (int i = 2; i <= m; i++)
 34         c[i] += c[i - 1];
 35     for (int i = n; i >= 1; i--)
 36         SA[c[x[i]]--] = i;
 37     for (int k = 1; k <= n; k <<= 1)
 38     
 39         int num = 0;
 40         for (int i = n - k + 1; i <= n; i++)    y[++num] = i;
 41         for (int i = 1; i <= n; i++)
 42             if (SA[i] > k)
 43                 y[++num] = SA[i] - k;
 44         for (int i = 1; i <= m; i++)
 45             c[i] = 0;
 46         for (int i = 1; i <= n; i++)
 47             c[x[i]]++;
 48         for (int i = 2; i <= m; i++)
 49             c[i] += c[i - 1];
 50         for (int i = n; i >= 1; i--)
 51             SA[c[x[y[i]]]--] = y[i], y[i] = 0;
 52         swap(x, y);
 53         x[SA[1]] = 1; num = 1;
 54         for (int i = 2; i <= n; i++)
 55             x[SA[i]] = (y[SA[i]] == y[SA[i - 1]] && y[SA[i] + k] == y[SA[i - 1] + k]) ? num : ++num;
 56         if (num == n) break;
 57         m = num;
 58     
 59     for (int i = 1; i <= n; i++)
 60         rak[SA[i]] = i;
 61     return;
 62 
 63 void get_height(int *s)
 64 
 65     int j, k = 0;
 66     for (int i = 1; i <= n; i++) 
 67         if (k) k--;
 68         j = SA[rak[i] - 1];
 69         while (i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
 70         height[rak[i]] = k;
 71     
 72 
//以上为板子
 73 int sta[MAXN];
 74 int top = 0;
//以下为核心
 75 bool check(int x)
 76 
 77     int cnt = 0;
 78     bool flag = 0;
 79     memset(vis, 0, sizeof(vis));
 80     for (int i = 2; i < n; i++)
 81     
 82         if (height[i] >= x)
 83         
 84             if (!vis[id[SA[i]]]) cnt++, vis[id[SA[i]]] = 1;
 85             if (!vis[id[SA[i - 1]]]) cnt++, vis[id[SA[i - 1]]] = 1;
 85.5　　　　　　 //查找到符合条件的子串
 86         
 87         else
 88         
 89             if (cnt > k)
 90             
 91                 if (!flag)
 91.5               //如果已经有解，清空 ans 
 92                     ans.clear();
 93                 ans.push_back(SA[i-1]);
 94                 flag = true;
 95             
 96             memset(vis, 0, sizeof(vis));
 97             cnt = 0;
 98         
 99     
102     return flag;
103 
104 int main()
105 
106     while(scanf("%d", &num))
107     
108         if (num == 0)
109             break;
110         down =100;
111         init();
112         ret.clear();
113         n = 0;
114         for (int i = 1; i <= num; i++)
115         
116             scanf("%s", tmp + 1);
117             int N = strlen(tmp+1);
118             for (int j = 1; j <= N; j++)
119             
120                 s[++n] = tmp[j] - ‘a‘ + 1;
121                 id[n] = i;
122             
123             s[++n] = ++down;
124             id[n] = -1;
125         
126         get_SA(s);
127         get_height(s);
128         int l = 1, r = n;
129         int len = 0;
130         k = num / 2;
131         while (l <= r)
132         
133         
134             int mid = (l + r) >> 1;    
135             if (check(mid))
136             
137                 len = max(len, mid);
138                 l = mid + 1;
139             
140             else
141             
142                 r = mid - 1;
143             
144         
145     
146         if    (len==1 || len==0)
147         
148             printf("?\n\n");
149             continue;
150         
151         for (int i = 0; i < ans.size(); i++)
152         
153             for (int j = ans[i]; j <= ans[i] + len-1; j++)
154                 printf("%c", s[j]+‘a‘-1);
155             printf("\n");
156         
157         printf("\n");
158     
159 
160     return 0;
161