POJ 1743 Musical Theme (二分後綴數組LCP)

題意

給出一段長度小於2e4的序列，如果有兩個不重疊子段的一一相鄰兩個數之間變化情況相同就可以說這兩段的是相同的theme。小於5視爲0。

思路

首先我們先對序列進行處理，a[i] = a[i+1] - a[i]，新序列即爲變化情況的數列，問題轉化成找到新序列的最長不重疊子段。
二分答案x然後可以利用lcp數組的性質找到所有符合條件的最左起點和最右起點，判斷和x的大小關係即可。

代碼

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define LL long long
#define Lowbit(x) ((x)&(-x))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1|1
#define MP(a, b) make_pair(a, b)
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int maxn = 1e5 + 10;
const double eps = 1e-8;
const double PI = acos(-1.0);
typedef pair<int, int> pii;

int n, k;
int sa[maxn*2], rk[maxn*2];
int temp[maxn*2], lcp[maxn*2];

int compare_sa(int i, int j)
{
    if (rk[i] != rk[j]) return rk[i] < rk[j];
    else
    {
        int ri = i + k <= n ? rk[i+k] : -1;
        int rj = j + k <= n ? rk[j+k] : -1;
        return ri < rj;
    }
}

void construct_sa(int *s, int *sa)
{
    for (int i = 0; i <= n; i++)
    {
        sa[i] = i;
        rk[i] = i < n ? s[i] : -1;
    }
    for (k = 1; k <= n; k *= 2)
    {
        sort(sa, sa + n + 1, compare_sa);
        temp[s[0]] = 0;
        for (int i = 1; i <= n; i++)
            temp[sa[i]] = temp[sa[i-1]] + (compare_sa(sa[i-1], sa[i]) ? 1 : 0);
        for (int i = 0; i <= n; i++)
            rk[i] = temp[i];
    }
}

void construct_lcp(int *s, int *sa, int *lcp)
{
    for (int i = 0; i <= n; i++) rk[sa[i]] = i;
    int h = 0;
    lcp[0] = 0;
    for (int i = 0; i < n; i++)
    {
        int j = sa[rk[i] - 1];
        if (h > 0) h--;
        for (; j + h < n && i + h < n; h++)
            if (s[j+h] != s[i+h]) break;
        lcp[rk[i] - 1] = h;
    }
}

bool check(int x)
{
    int mmin = INF, mmax = -INF;
    for (int i = 0; i < n; i++)
        if (lcp[i] >= x)
        {
            mmin = min(mmin, min(sa[i], sa[i+1]));
            mmax = max(mmax, max(sa[i], sa[i+1]));
            if (mmax - mmin > x) return true;
        }
        else
        {
            mmin = INF;
            mmax = -INF;
        }
    return false;
}

void solve()
{
    int l = 0, r = n / 2;
    while (l < r)
    {
        int mid = (l + r + 1) >> 1;
        if (check(mid)) l = mid;
        else r = mid - 1;
    }
    l++;
    if (l < 5) l = 0;
    printf("%d\n", l);
}

int a[maxn];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while (scanf("%d", &n) && n)
    {
        for (int i = 0; i < n; i++)
            scanf("%d", &a[i]);
        n--;                         //原序列變形式
        for (int i = 0; i < n; i++)
            a[i] = a[i+1] - a[i] + 100;
        construct_sa(a, sa);
        construct_lcp(a, sa, lcp);
        solve();
    }
    return 0;
}