【POJ 3693】Maximum repetition substring（后缀数组）

Maximum repetition substring
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 13195   Accepted: 4077

Description

The repetition number of a string is defined as the maximum number R such that the string can be partitioned into R same consecutive substrings. For example, the repetition number of "ababab" is 3 and "ababa" is 1.

Given a string containing lowercase letters, you are to find a substring of it with maximum repetition number.

Input

The input consists of multiple test cases. Each test case contains exactly one line, which
gives a non-empty string consisting of lowercase letters. The length of the string will not be greater than 100,000.

The last test case is followed by a line containing a '#'.

Output

For each test case, print a line containing the test case number( beginning with 1) followed by the substring of maximum repetition number. If there are multiple substrings of maximum repetition number, print the lexicographically smallest one.

Sample Input

ccabababc
daabbccaa
#

Sample Output

Case 1: ababab
Case 2: aa
 
 

【题目大意】

找到 n 个子串中，重复次数最多的字符串，如果重复次数相同，输出字典序最小的那个 

 

【题目分析】

 首先明确题意，要找重复次数最多的子串

然后考虑枚举这个子串循环节的长度

可以想到，第s [ i ] 和 第 s [ i + l ] 如果重复的话后面是有相同的子串的

而我们求以他们为首的两个后缀的 LCP，因为他们直接 那 i 个字符的影响，使后面能够找到适合的，延伸的所要求的子串

而这个子串的长度，除以我们当前枚举的循环节的长度，就是这个循环节重复的次数（这不是显而易见的嘛）

 然后因为这 i 个字符，并不一定是循环节，所以要再利用 LCP 和长度计算出，真实的循环子串开始的位置

然后记录

 

之后，因为SA数组天然是按字典序排的，所以我们直接在SA[ i ] 上

找合适的子串即可

 

 

 

【代码】

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<string>
  4 #include<iostream>
  5 #include<algorithm>
  6 using namespace std;
  7 const int MAX_CHAR = 128;
  8 const int MAXN     = 100000 + 5;
  9 const int INF      = 2e9 + 7;
 10 int lg[MAXN], f[MAXN], g[MAXN];
 11 struct suff_string
 12 {
 13     int n, m;
 14     int x[MAXN] , y[MAXN] , c[MAXN] ;
 15     int SA[MAXN], rak[MAXN], height[MAXN];
 16     int st[17][MAXN];
 17     void build_SA(char *s)
 18     {
 19         n = strlen(s + 1);
 20         m = MAX_CHAR;
 21         for (int i = 1; i <= 100000; i++)
 22             x[i] = y[i] = c[i] = SA[i]=rak[i]=height[i]=0;
 23         for (int i = 1; i <= n; i++)
 24             SA[i] = rak[i] = height[i] = 0;
 25         for (int i = 1; i <= n; i++)
 26             ++c[x[i] = s[i]];
 27         for (int i = 2; i <= m; i++)
 28             c[i] += c[i - 1];
 29         for (int i = n; i >= 1; i--)
 30             SA[c[x[i]]--] = i;
 31         for (int k = 1; k <= n; k <<= 1)
 32         {
 33             int num = 0;
 34             for (int i = n - k + 1; i <= n; i++)    y[++num] = i;
 35             for (int i = 1; i <= n; i++)
 36                 if (SA[i] > k)
 37                     y[++num] = SA[i] - k;
 38             for (int i = 1; i <= m; i++)
 39                 c[i] = 0;
 40             for (int i = 1; i <= n; i++)
 41                 c[x[i]]++;
 42             for (int i = 2; i <= m; i++)
 43                 c[i] += c[i - 1];
 44             for (int i = n; i >= 1; i--)
 45                 SA[c[x[y[i]]]--] = y[i], y[i] = 0;
 46             swap(x, y);
 47             x[SA[1]] = 1; num = 1;
 48             for (int i = 2; i <= n; i++)
 49                 x[SA[i]] = (y[SA[i]] == y[SA[i - 1]] && y[SA[i] + k] == y[SA[i - 1] + k]) ? num : ++num;
 50             if (num == n) break;
 51             m = num;
 52         }
 53         for (int i = 1; i <= n; i++)
 54             rak[SA[i]] = i;
 55         return;
 56     }
 57     void get_height(char *s)
 58     {
 59         int k = 0;
 60         for (int i = 1; i <= n; i++) {
 61             if (k) k--;
 62             int j = SA[rak[i] - 1];
 63             while (i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
 64             height[rak[i]] = k;
 65         }
 66     }
 67 
 68     void build_ST()
 69     {
 70         memset(st, 0, sizeof(st));
 71         for (int i = 1; i <= n; i++)
 72             st[0][i] = height[i];
 73         for (int i = 1; i <= 16; i++)
 74             for (int j = 1; j <= n; j++)
 75                 st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
 76     }
 77     int query(int l, int r)
 78     {
 79         l = rak[l], r = rak[r];
 80         if (l > r)    swap(l, r); l++;
 81         int t = lg[r + 1 - l];
 82         return min(st[t][l], st[t][r - (1 << t) + 1]);
 83     }
 84 }A;
 85 char s[MAXN];
 86 int _ = 0;
 87 int ans[MAXN];
 88 bool solve()
 89 {
 90     scanf("%s", s + 1);
 91     if (s[1] == '#')
 92         return 0;
 93     A.build_SA(s);  A.get_height(s);  A.build_ST();
 94     int n = A.n;
 95     int mx = -1;
 96     int len = 0;
 97     for (int i = 1; i < n; i++)
 98     {
 99         for (int j = 1; j+i <= n; j += i)
100         {
101             int k = A.query(j,j+i);
102             int t = k / i +1;
103             int left = i - (k%i);
104             int head =j- left;
105             if (head >= 1 && A.query(head, head + i)>=left) t++;
106             if (t > mx)
107             {
108                 mx = t;
109                 ans[0] = 0;
110             }
111             if (t == mx)
112             {
113                 ans[++ans[0]] = i;
114             }
115         }
116     }
117     int head=INF;
118     int flag = 0;
119     for (int i = 1; i <= n; i++)
120     {
121         for (int j = 1; j <= ans[0]; j++)
122         {
123             int l = ans[j];
124             if (A.query(A.SA[i], A.SA[i] + l) >= (mx - 1)*l)
125             {
126                 len = l;
127                 head = A.SA[i];
128                 flag = 1;
129                 break;
130             }
131         }
132         if (flag)    break;
133     }
134     printf("Case %d: ", ++_);
135     for (int i = 0; i <mx*len; i++)
136         printf("%c", s[head+i]);
137     printf("\n");
138     return 1;
139 }
140 int main()
141 {
142     int T;
143     for (int i = 2; i <= 100000; i++) lg[i] = lg[i >> 1] + 1;
144     scanf("%d", &T);
145     getchar();
146     while (T--  && solve())
147     {
148 
149     }
150     return 0;
151 }

 

posted @ 2019-10-25 15:31  rentu  阅读(148)  评论(0)  编辑  收藏  举报