POJ - 1743 Musical Theme (後綴數組求不可重疊最長重復子串)

POJ - 1743 Musical Theme (後綴數組求不可重疊最長重復子串)

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

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

題意：有N(1 <= N <=20000)個音符的序列來表示一首樂曲，每個音符都是1..88范圍內的整數，現在要找一個重復的主題。

• “主題”是整個音符序列的一個子串，它需要滿足如下條件：
• 1.長度至少為5個音符
• 2.在樂曲中重復出現(可能經過轉調，“轉調”的意思是主題序列中每個音符都被加上或減去了同一個整數值。)
• 3.重復出現的同一主題不能有公共部分。

  思路：第一道後綴數組啊，無力的看了一天的資料，學著別人敲了一遍，先轉化成相鄰兩項的差值，然後就是找不可重疊重復子串。

  • 看了論文的做法是：。做法是二分，將題目變成如下判定性問題：是否存在兩個長度為 k 的不重疊子串完全相同。
    然後此問題的做法是，將排序後的後綴分成若干組，每組相鄰後綴之間的 height[] 都不小於k。
    直觀上看，這樣分組等價於使每組的這些後綴都擁有某個長度不小於 k 的共同的前綴，即重復子串。再判斷下是否能滿足不重疊的條件即可。

    #include 
    #include 
    #include 
    #include 
    #include 
    using namespace std;
    const int maxn = 20010;
    
    int sa[maxn]; //SA數組，表示將S的n個後綴從小到大排序後把排好序的
    //的後綴的開頭位置順次放入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;
    }