后缀数组入门+poj1743

题目链接：Musical Theme

题意:给一个数组表示一段音乐，范围1~88；问为最长的的位置不相交的相似的旋律。相似的意思是，整段加上某个值，比如1 2 3 4 5 和 6 7 8 9 10是相似。

题解：求出后缀数组，高度数组lcp，枚举答案x ，找到一个对于连续的lcp[i]>=x,找到看最小的sa[i],最大的sa[i]，看是否差值大于l,就可以判断是否重叠。

//#include<bits/stdc++.h>
#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define pb push_back
#define ll long long
#define PI 3.14159265
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define ws wppp
#define eps 1e-7
using namespace std;
const int N=2e4+5;
const int mod=1e9+7;
int a[N], s[N];
int sa[N], t[N], t2[N], c[N], n;
int ran[N], height[N];
const int inf=0x3fffffff;
void get_sa(int m)
{
    int i, *x = t, *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(int k = 1; k <= n; k <<= 1)
    {
        int p = 0;
        for(i = n-k; i < n; i++) y[p++] = i;
        for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
        for(i = 0; i < m; i++) c[i] = 0;
        for(i = 0; i < n; i++) c[x[y[i]]]++;
        for(i = 0; 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]+k] == y[sa[i]+k] ? p-1 : p++;
        if(p >= n) break;
        m = p;
    }
    int k = 0;
    for(i = 0; i < n; i++) ran[sa[i]] = i;
    for(i = 0; i < n; i++)
    {
        if(k) k--;
        int j = sa[ran[i]-1];
        while(s[i+k] == s[j+k]) k++;
         height[ran[i]] = k;
    }
}
bool judge(int x)
{
    int mi=inf,mx=-inf;
    for(int i=1;i<n;i++)
    {
        if(height[i]>=x)
        {
            mi=min(mi,min(sa[i],sa[i-1]));
            mx=max(mx,max(sa[i],sa[i-1]));
            if(mx-mi>x)return true;
        }
        else
        {
            mi=inf,mx=-inf;
        }
    }
    return false;
}
int main()
{
    while(scanf("%d",&n)&&n)
    {
        for(int i=0;i<n;i++)scanf("%d",&a[i]);
        for(int i=0;i<n-1;i++)
        {
            s[i]=a[i+1]-a[i]+100;
        }
        s[n-1]=0;
         get_sa(200);
        int l=0,r=n;
        while(l<=r)
        {
            int mid=l+r>>1;
            if(judge(mid))l=mid+1;
            else r=mid-1;
        }
        if(r>=4)printf("%d
",r+1);
        else printf("0
");
    }
    return 0;
}