POJ - 1743 Musical Theme (后缀数组求不可重叠最长重复子串)

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

3025 27 30 34 39 45 52 60 69 79 69 60 52 45 39 34 30 26 22 1882 78 74 70 66 67 64 60 65 800

Sample Output

5

题意：有N(1 <= N <=20000)个音符的序列来表示一首乐曲，每个音符都是1..88范围内的整数，现在要找一个重复的主题。

• “主题”是整个音符序列的一个子串，它需要满足如下条件：
• 1.长度至少为5个音符
• 2.在乐曲中重复出现(可能经过转调，“转调”的意思是主题序列中每个音符都被加上或减去了同一个整数值。)
• 3.重复出现的同一主题不能有公共部分。

思路：第一道后缀数组啊，无力的看了一天的资料，学着别人敲了一遍，先转化成相邻两项的差值，然后就是找不可重叠重复子串。

• 看了论文的做法是：。做法是二分，将题目变成如下判定性问题：是否存在两个长度为 k 的不重叠子串完全相同。
  然后此问题的做法是，将排序后的后缀分成若干组，每组相邻后缀之间的 height[] 都不小于k。
  直观上看，这样分组等价于使每组的这些后缀都拥有某个长度不小于 k 的共同的前缀，即重复子串。再判断下是否能满足不重叠的条件即可。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int maxn = 20010;int sa[maxn]; //SA数组，表示将S的n个后缀从小到大排序后把排好序的//的后缀的开头位置顺次放入SA中int t1[maxn], t2[maxn], c[maxn];int rank[maxn], height[maxn];int s[maxn];void build_sa(int s[], int n, int m) {    int i, j, p, *x = t1, *y = t2;    for (i = 0; i < m; i++) c[i] = 0;    for (i = 0; i < n; i++) c[x[i] = s[i]]++;    for (i = 1; i < m; i++) c[i] += c[i-1];    for (i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;    for (j = 1; j <= n; j <<= 1) {        p = 0;        for (i = n-j; i < n; i++) y[p++] = i;        for (i = 0; i < n; i++)             if (sa[i] >= j)                 y[p++] = sa[i] - j;        for (i = 0; i < m; i++) c[i] = 0;        for (i = 0; i < n; i++) c[x[y[i]]]++;        for (i = 1; i < m; i++) c[i] += c[i-1];        for (i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];        swap(x, y);        p = 1, x[sa[0]] = 0;        for (i = 1; i < n; i++)             x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+j] == y[sa[i]+j] ? p-1 : p++;        if (p >= n) break;        m = p;    }}void getHeight(int s[],int n) {    int i, j, k = 0;    for (i = 0; i <= n; i++)        rank[sa[i]] = i;    for (i = 0; i < n; i++) {        if (k) k--;        j = sa[rank[i]-1];        while (s[i+k] == s[j+k]) k++;        height[rank[i]]=k;    }}int check(int n,int k) {    int Max = sa[1], Min = sa[1];    for (int i = 2; i <= n; i++) {        if (height[i] < k)            Max = Min = sa[i];        else {            if (sa[i] < Min) Min = sa[i];            if (sa[i] > Max) Max = sa[i];            if (Max - Min > k) return 1;        }    }    return 0;}int main() {    int n;    while (scanf("%d", &n) != EOF && n) {        for (int i = 0; i < n; i++)            scanf("%d", &s[i]);        for (int i = n-1; i > 0; i--)            s[i] = s[i] - s[i-1] + 90;        n--;        for (int i = 0; i < n; i++)            s[i] = s[i+1];        s[n] = 0;        build_sa(s, n+1, 200);        getHeight(s, n);        int ans = -1;        int l = 1, r = n/2;        while (l <= r) {            int mid = l + r >> 1;            if (check(n, mid)) {                ans = mid;                l = mid + 1;            }            else r = mid - 1;        }        if (ans < 4)            printf("0\n");        else printf("%d\n", ans+1);    }    return 0;}