地址：http://poj.org/problem?id=1743
题目：
Musical Theme
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 28675 | Accepted: 9674 |
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.
Sample Input
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
Sample Output
5
Hint
Use scanf instead of cin to reduce the read time.
Source
思路：
论文上的题目，相邻数做差后对数组求后缀数组（注意此时差可能为负数，所以加上100保证非负）
然后二分长度，check时把height分组判断即可
1 #include <cstdlib>
2 #include <cstring>
3 #include <cstdio>
4 #include <algorithm>
5
6 const int N = 20005;
7 int sa[N],s[N],wa[N], wb[N], ws[N], wv[N];
8 int rank[N], height[N];
9
10 bool cmp(int r[], int a, int b, int l)
11 {
12 return r[a] == r[b] && r[a+l] == r[b+l];
13 }
14
15 void da(int r[], int sa[], int n, int m)
16 {
17 int i, j, p, *x = wa, *y = wb;
18 for (i = 0; i < m; ++i) ws[i] = 0;
19 for (i = 0; i < n; ++i) ws[x[i]=r[i]]++;
20 for (i = 1; i < m; ++i) ws[i] += ws[i-1];
21 for (i = n-1; i >= 0; --i) sa[--ws[x[i]]] = i;
22 for (j = 1, p = 1; p < n; j *= 2, m = p)
23 {
24 for (p = 0, i = n - j; i < n; ++i) y[p++] = i;
25 for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
26 for (i = 0; i < n; ++i) wv[i] = x[y[i]];
27 for (i = 0; i < m; ++i) ws[i] = 0;
28 for (i = 0; i < n; ++i) ws[wv[i]]++;
29 for (i = 1; i < m; ++i) ws[i] += ws[i-1];
30 for (i = n-1; i >= 0; --i) sa[--ws[wv[i]]] = y[i];
31 for (std::swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; ++i)
32 x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++;
33 }
34 }
35
36 void calheight(int r[], int sa[], int n)
37 {
38 int i, j, k = 0;
39 for (i = 1; i <= n; ++i) rank[sa[i]] = i;
40 for (i = 0; i < n; height[rank[i++]] = k)
41 for (k?k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
42 }
43 bool check(int len,int x)
44 {
45 int mx=sa[1],mi=sa[1];
46 for(int i=2;i<=len;i++)
47 {
48 if(height[i]<x)
49 {
50 if(mx-mi>=x)
51 return 1;
52 mx=mi=sa[i];
53 }
54 else
55 {
56 mx=std::max(sa[i],mx);
57 mi=std::min(sa[i],mi);
58 }
59 }
60 return mx-mi>=x;
61 }
62 int main()
63 {
64 int n;
65 while(scanf("%d",&n)&&n)
66 {
67 int x,ls,len=0;
68 scanf("%d",&ls);
69 for(int i=0;i<n-1;i++)
70 scanf("%d",&x),s[len++]=x-ls+100,ls=x;
71 s[len]=0;
72 da(s,sa,len+1,200);
73 calheight(s,sa,len);
74 int l=5,r=n/2,ans=0;
75 while(l<=r)
76 {
77 int mid=l+r>>1;
78 if(check(len,mid-1))
79 ans=mid,l=mid+1;
80 else
81 r=mid-1;
82 }
83 printf("%d\n",ans);
84 }
85 return 0;
86 }