POJ 1743后缀数组+二分

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

有n个数，求不重叠的最长相同变化的子串，小于5的话就输出0。
比如1,2,3,4,5,6,7,8,9,10  1,2,3,4,5和6,7,8,9,10是都数变化为1且不重叠的两个，所以答案是5 。而对于1,2,3,4,5,6,7,8,9  最多只有1,2,3,4和5,6,7,8相同且不重叠所以答案是0。
可以用后缀数组中的 height数组可以求出来。
height[i]:表示前缀排名为i和前缀排名为i-1的最大公共前缀。
第56行必须是L+m+r <= R，只有这样成立才不会出现重叠。

#include <iostream>
#include <string.h>
#include <stdio.h>
#define INF 0x3f3f3f3f
using namespace std;
const int MAX = 2e4+10;
int *ran, r[MAX], sa[MAX], height[MAX], wa[MAX], wb[MAX], wm[MAX];
bool cmp(int *r, int a, int b, int l) {
    return r[a]==r[b]&&r[a+l]==r[b+l];
}
void da(int *r, int *sa, int n, int m) {
    int *x = wa, *y = wb, *t;
    for(int i = 0; i < m; i ++) wm[i] = 0;
    for(int i = 0; i < n; i ++) wm[x[i]=r[i]]++;
    for(int i = 1; i < m; i ++) wm[i] += wm[i-1];
    for(int i = n-1; i >= 0; i --) sa[--wm[x[i]]] = i;
    for(int i = 0, j = 1, p = 0; p < n; j<<=1,m = p) {
        for(p = 0, i = n-j; i < n; i ++) y[p++] = i;
        for(int i = 0; i < n; i ++) if(sa[i] >= j) y[p++] = sa[i] - j;
        for(int i = 0; i < m; i ++) wm[i] = 0;
        for(int i = 0; i < n; i ++) wm[x[y[i]]]++;
        for(int i = 0; i < m; i ++) wm[i] += wm[i-1];
        for(int i = n-1; i >= 0; i --) sa[--wm[x[y[i]]]] = y[i];
        for(t=x,x=y,y=t,i=p=1,x[sa[0]]=0; i < n; ++i) {
            x[sa[i]] = cmp(y,sa[i],sa[i-1],j)?p-1:p++;
        }
    }
    ran = x;
}
void calheight(int *r, int *sa, int n) {
    for(int i=0,j=0,k=0;i < n; height[ran[i++]]=k){
        for(k?--k:0,j=sa[ran[i]-1];r[i+k] == r[j+k];++k) {
        }
    }
}
int main() {
    int n;
    while(scanf("%d",&n)&&n) {
        for(int i = 0; i < n; i ++) scanf("%d",&r[i]);
        for(int i = 0; i < n-1; i ++) r[i] = r[i+1] - r[i] + 100;
        r[--n] = 0;
        da(r,sa,n+1,200);
        calheight(r,sa,n);
        int l = 1, r = n+1, m, L, R;
        bool flag = false;
        while(l <= r) {
            m = (l+r)>>1;
            L = INF, R = -INF,flag=false;
            for(int i = 1; i <= n; i ++) {
                if(height[i] >= m) {
                    L = min(L,sa[i]);
                    L = min(L,sa[i-1]);
                    R = max(R,sa[i]);
                    R = max(R,sa[i-1]);
                }else {
                    if(L+m+1 <= R) flag = true;
                    L = INF, R = -INF;
                }
            }
            if(L+m+1 <= R) flag = true;
            if(flag) l = m+1;
            else r = m-1;
        }
        if(l >= 5)printf("%d
",l);
        else printf("0
");
    }
    return 0;
}