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题目概述：
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.
给你一段乐章，乐章由n个音符组成，这些音符由1..88的整数表示。现在你需要找出最长的一段旋律，使得这一段旋律至少有5个音符，至少出现过两次，且这两次的位置不相交（如果两段旋律经过变调后是一样的，那么这两段旋律可以看成出现过两次）。
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输入格式：
输入包含多组数据，每组数据的第一行包括一个整数n，表示乐章的长度。接下来一行包括n个整数，表示每个音符。
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输出格式：
输出一个整数，表示满足题意的最长的旋律的长度。
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样例输入：
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
0
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样例输出：
5
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思路：
表示...字符串的题真心鬼畜。半天%完罗穗骞的《后缀数组——处理字符串的有力工具》后，彻底滚粗（kmp，心中永远的痛）。
先做完了UOJ#35的模板题，又调了半天，内心是崩溃的。
于是开始lg这道题。
一道后缀自动机的模板题。用相邻两个数的差构造新的串，对新串用后缀数组lg一下，然后二分一下求不相交子串的最大长度。
注意n=1的情况（可以和n<10的情况一起判了）。
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 using namespace std;
5 const int MAXN = 20005;
6 int s[MAXN], sx[MAXN], sy[MAXN], ssum[MAXN], sa[MAXN], rk[MAXN], h[MAXN];
7 inline bool cmp(int* y, int i, int j, int k) {
8 return y[i] == y[j] && y[i + k] == y[j + k];
9 }
10
11 void getsa(int n, int m) {
12 int i, j, p, *x = sx, *y = sy;
13 for(i = 0; i < m; ++i) ssum[i] = 0;
14 for(i = 0; i < n; ++i) ++ssum[x[i] = s[i]];
15 for(i = 1; i < m; ++i) ssum[i] += ssum[i - 1];
16 for(i = n - 1; ~i; --i) sa[--ssum[x[i]]] = i;
17 for(p = j = 1; p < n; j <<= 1, m = p) {
18 for(p = 0, i = n - j; i < n; ++i) y[p++] = i;
19 for(i = 0; i < n; ++i) if(sa[i] >= j) y[p++] = sa[i] - j;
20 for(i = 0; i < m; ++i) ssum[i] = 0;
21 for(i = 0; i < n; ++i) ++ssum[x[y[i]]];
22 for(i = 1; i < m; ++i) ssum[i] += ssum[i - 1];
23 for(i = n - 1; ~i; --i) sa[--ssum[x[y[i]]]] = y[i];
24 swap(x, y); x[sa[0]] = 0;
25 for(p = i = 1; i < n; ++i) x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
26 }
27 }
28
29 void geth(int n) {
30 int j, k = 0;
31 for(int i = 1; i <= n; ++i) rk[sa[i]] = i;
32 for(int i = 0; i < n; h[rk[i++]] = k)
33 for(k ? k-- : 0, j = sa[rk[i] - 1]; s[i + k] == s[j + k]; ++k);
34 }
35
36 int solve(int x, int n) {
37 int i = 2, minh, maxh;
38 while(1) {
39 while(i <= n && h[i] < x) ++i;
40 if(i > n) break;
41 maxh = minh = sa[i - 1];
42 while(i <= n && h[i] >= x) {
43 minh = min(minh, sa[i]); maxh = max(maxh, sa[i++]);
44 }
45 if(maxh - minh > x) return true;
46 }
47 return false;
48 }
49
50 int main() {
51 int n;
52 while(~scanf("%d", &n) && n) {
53 memset(sx, 0, sizeof(sx));
54 memset(sy, 0, sizeof(sy));
55 memset(sx, 0, sizeof(sa));
56 memset(ssum, 0, sizeof(ssum));
57 for(int i = 0; i < n; ++i) scanf("%d", s + i);
58 if(--n < 9) { puts("0"); continue; }
59 for(int i = 0; i < n; ++i) s[i] = s[i + 1] - s[i] + 90;
60 s[n] = 0;
61 getsa(n + 1, 200); geth(n);
62 int l = 1, r = (n >> 1) + 1;
63 while(l < r - 1) {
64 int mid = l + r >> 1;
65 if(solve(mid, n)) l = mid;
66 else r = mid;
67 }
68 printf("%d\n", l < 4 ? 0 : r);
69 }
70 return 0;
71 }
（话说论文里的变量名好奇怪）
抱歉博客更晚了，以后应该可以按时更新。