Brian Dean
A musical melody is represented as a sequence of N (1 <= N <= 5000) 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 "theme", 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!
PROGRAM NAME: theme
INPUT FORMAT
The first line of the input file contains the integer N. Each subsequent line (except potentially the last) contains 20 integers representing the sequence of notes. The last line contains the remainder of the notes, potentially fewer than 20.
SAMPLE INPUT (file theme.in)
30 25 27 30 34 39 45 52 60 69 79 69 60 52 45 39 34 30 26 22 18 82 78 74 70 66 67 64 60 65 80
OUTPUT FORMAT
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.
SAMPLE OUTPUT (file theme.out)
5
[The five-long theme is the last five notes of the first line and the first five notes of the second]
ANALYSIS
At first,我们可以知道"Transposed"是可以用相邻两数之差表示。
FOR EXAMPLE:
notes:1,2,3,4,5,6,7,8,9,10
change into:1,1,1,1,1,1,1,1,1,1,-10 (PAY ATTENTION:这里有11个元素，我们把上面的序列前面和后面增加一个0)
然后，枚举开头O(N)，检查O(N)。所以就是平方级算法了，可以通过此题。
检查可以用KMP的思想（其实我就是用了KMP）。
/* TASK:theme LANG:C++ */ #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int n, a[5005], f[5005], ans; void read() { int before, x; scanf("%d", &n); before = 0; for (int i = 0; i < n; ++i) { scanf("%d", &x); a[i] = x - before; before = x; } a[n] = -before; } int main() { freopen("theme.in", "r", stdin); freopen("theme.out", "w", stdout); read(); ans = 0; for (int fr = n - 3; fr >= 5; --fr) { memset(f, 0, sizeof(f)); f[0] = f[1] = 0; for (int i = fr + 1; i <= n; ++i) { int j = f[i - fr]; while (j != 0 && a[i] != a[fr + j]) j = f[j]; f[i - fr + 1] = (a[i] == a[fr + j]) ? j + 1 : 0; } int j = 0; for (int i = 0; i < fr - 1; ++i) { while (j != 0 && a[i] != a[fr + j]) j = f[j]; if (a[i] == a[fr + j]) { j++; ans = max(ans, j + 1); } } } if (ans < 5) printf("0 "); else printf("%d ", ans); return 0; }