POJ 1743 Musical Theme（后缀数组）

Description

A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the piano. It is unfortunate but true that this representation of melodies ignores the notion of musical timing; but, this programming task is about notes and not timings. 
Many composers structure their music around a repeating &qout;theme&qout;, which, being a subsequence of an entire melody, is a sequence of integers in our representation. A subsequence of a melody is a theme if it: 

• is at least five notes long 
• appears (potentially transposed -- see below) again somewhere else in the piece of music 
• is disjoint from (i.e., non-overlapping with) at least one of its other appearance(s)

Transposed means that a constant positive or negative value is added to every note value in the theme subsequence. 
Given a melody, compute the length (number of notes) of the longest theme. 
One second time limit for this problem's solutions! 

Input

The input contains several test cases. The first line of each test case contains the integer N. The following n integers represent the sequence of notes. 
The last test case is followed by one zero. 

Output

For each test case, the output file should contain a single line with a single integer that represents the length of the longest theme. If there are no themes, output 0.

 

题目大意：给N个数字，对着N个数字前后作差得到一段N-1个数字的序列，问重复出现两次且不重叠的子段最长有多长。（看清楚上面的题目噢）

思路：N个数字作差，得到N-1个数字的序列，求后缀数组，二分验证是否存在mid长度的符合要求的字符串。

 

代码（219MS）：

 1 #include <cstdio>

 2 #include <algorithm>

 3 #include <iostream>

 4 #include <cstring>

 5 using namespace std;

 6 

 7 const int MAXN = 20010;

 8 int s[MAXN];

 9 int n, sa[MAXN], height[MAXN], rank[MAXN], tmp[MAXN], c[MAXN];

10 

11 void makesa(int m) {

12     memset(c, 0, sizeof(c));

13     for(int i = 0; i < n; ++i) ++c[rank[i] = s[i]];

14     for(int i = 0; i < m; ++i) c[i] += c[i - 1];

15     for(int i = 0; i < n; ++i) sa[--c[rank[i]]] = i;

16     for(int k = 1; k < n; k <<= 1) {

17         for(int i = 0; i < n; ++i) {

18             int j = sa[i] - k;

19             if(j < 0) j += n;

20             tmp[c[rank[j]]++] = j;

21         }

22         int j = c[0] = sa[tmp[0]] = 0;

23         for(int i = 1; i < n; ++i) {

24             if(rank[tmp[i]] != rank[tmp[i - 1]] || rank[tmp[i] + k] != rank[tmp[i - 1] + k])

25                 c[++j] = i;

26             sa[tmp[i]] = j;

27         }

28         memcpy(rank, sa, n * sizeof(int));

29         memcpy(sa, tmp, n * sizeof(int));

30         if(j >= n - 1) break;

31     }

32 }

33 

34 void calheight() {

35     int j, k = 0;

36     for(int i = 0; i < n; height[rank[i++]] = k) {

37         if(k > 0) --k;

38         for(j = sa[rank[i] - 1]; s[i + k] == s[j + k]; ++k);

39     }

40 }

41 

42 bool check(int k) {

43     int mx = sa[0], mn = sa[0];

44     for(int i = 1; i < n; ++i) {

45         if(height[i] >= k - 1) {

46             mx = max(mx, sa[i]);

47             mn = min(mn, sa[i]);

48         } else {

49             if(mx - mn >= k) return true;

50             mx = mn = sa[i];

51         }

52     }

53     return mx - mn >= k;

54 }

55 

56 int solve() {

57     int l = 1, r = n;

58     while(l < r) {

59         int mid = (l + r) >> 1;

60         if(check(mid)) l = mid + 1;

61         else r = mid;

62     }

63     return l - 1;

64 }

65 

66 int main() {

67     while(scanf("%d", &n) != EOF && n) {

68         for(int i = 0; i < n; ++i) scanf("%d", &s[i]);

69         for(int i = 0; i < n - 1; ++i) s[i] = s[i + 1] - s[i] + 88;

70         s[n - 1] = 0;

71         makesa(176);

72         if(n != 1) calheight();

73         int ans = solve();

74         printf("%d\n", ans >= 5 ? ans : 0);

75     }

76 }

View Code