POJ 1743 Musical Theme 后缀数组最长重复不相交子串

Musical Theme
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://poj.org/problem?id=1743

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

HINT

题意

让你找到最长的不相交重复子串

题解：

后缀数组，跑出height数组之后，然后直接二分然后O(n)，check就好了

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 25000
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

int s[2*maxn];
int a[maxn];
int n;
int sa[maxn], rank[maxn], height[maxn];
int wa[maxn], wb[maxn], wv[maxn], wd[maxn];

int cmp(int *r, int a, int b, int l){
    return r[a] == r[b] && r[a+l] == r[b+l];
}

void build_sa(int *r, int n, int m){          //  倍增算法 r为待匹配数组  n为总长度 m为字符范围
    int i, j, p, *x = wa, *y = wb, *t;
    for(i = 0; i < m; i ++) wd[i] = 0;
    for(i = 0; i < n; i ++) wd[x[i]=r[i]] ++;
    for(i = 1; i < m; i ++) wd[i] += wd[i-1];
    for(i = n-1; i >= 0; i --) sa[-- wd[x[i]]] = i;
    for(j = 1, p = 1; p < n; j *= 2, m = p){
        for(p = 0, 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 < n; i ++) wv[i] = x[y[i]];
        for(i = 0; i < m; i ++) wd[i] = 0;
        for(i = 0; i < n; i ++) wd[wv[i]] ++;
        for(i = 1; i < m; i ++) wd[i] += wd[i-1];
        for(i = n-1; i >= 0; i --) sa[-- wd[wv[i]]] = y[i];
        for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i ++){
            x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p - 1: p ++;
        }
    }
}

void calHeight(int *r, int n){           //  求height数组。
    int i, j, k = 0;
    for(i = 1; i <= n; i ++) rank[sa[i]] = i;
    for(i = 0; i < n; height[rank[i ++]] = k){
        for(k ? k -- : 0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k ++);
    }
}
int check(int mid)
{
    int Minn=sa[0],Maxn=sa[0];
    for(int i=0;i<=n;i++)
    {
        if(height[i]>=mid)
        {
            Minn=min(sa[i],Minn);
            Maxn=max(sa[i],Maxn);
            if(Maxn-Minn>mid)
                return 1;
        }
        else
            Minn=Maxn=sa[i];
    }
    return 0;
}
void init()
{
    memset(a,0,sizeof(a));
    memset(s,0,sizeof(s));
    memset(sa,0,sizeof(sa));
    memset(rank,0,sizeof(rank));
    memset(height,0,sizeof(height));
    memset(wa,0,sizeof(wa));
    memset(wb,0,sizeof(wb));
    memset(wv,0,sizeof(wv));
    memset(wd,0,sizeof(wd));
}
int main()
{
    while(cin>>n)
    {
        if(n==0)
            break;
        init();
        for(int i=0;i<n;i++)
            a[i]=read();
        n--;
        for(int i=0;i<n;i++)
            s[i]=a[i+1]-a[i]+100;
        s[n]=0;
        build_sa(s,n+1,200);
        calHeight(s,n);
        int l=0,r=n,ans=0;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            if(check(mid))
            {
                ans=max(ans,mid);
                l=mid+1;
            }
            else
                r=mid-1;
        }
        if(ans<4)
            ans=0;
        else
            ans++;
        cout<<ans<<endl;
    }
}

POJ 1743 Musical Theme 后缀数组 最长重复不相交子串

POJ 1743 Musical Theme 后缀数组最长重复不相交子串