poj1743 Musical Theme（后缀数组|后缀自动机）

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Musical Theme

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 22953 Accepted: 7839

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

3025 27 30 34 39 45 52 60 69 79 69 60 52 45 39 34 30 26 22 1882 78 74 70 66 67 64 60 65 800

Sample Output

Hint

Use scanf instead of cin to reduce the read time.

Source

LouTiancheng@POJ

2015-11-27

【思路】

不可重叠最长重复子串。

　　1) 据题意处理字符串

　　2) 后缀数组。二分长度k，问题成为了判定是否存在两个及以上长度不小于k且互不重叠的子串。根据height数组划分后缀，满足两个条件：一是一组内height值不小于k（保证组内任两个长度不小于k即存在长度不小于k的子串），二是组内后缀sa值的最大最小值之差大于等于k（保证两个子串不重叠）。

　　需要注意n==1时需要特判。

　　1/为什么可以划分height数组呢？首先height[i]代表lcp(suffix(sa[i]),suffix(sa[i-1]))，所以height所对应的后缀是有序的，如果划分出现height<k的话以后的后缀与改组内的lcp一定不大于k-1，所以不会出现后面的再划分到改组的情况。

【代码】

 1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #define FOR(a,b,c) for(int a=(b);a<=(c);a++) 5 using namespace std; 6  7 const int maxn = 40000+10; 8  9 int s[maxn];10 int sa[maxn],t[maxn],t2[maxn],c[maxn],n;11 //字符串s长度为n12 //build_sa中参数m表示字符最大为m13 void build_sa(int m) {14     int i,*x=t,*y=t2;15     for(int i=0;i<m;i++) c[i]=0;16     for(int i=0;i<n;i++) c[x[i]=s[i]]++;17     for(int i=1;i<m;i++) c[i]+=c[i-1];18     for(int i=n-1;i>=0;i--) sa[--c[x[i]]]=i;19     20     for(int k=1;k<=n;k<<=1) {21         int p=0;22         for(int i=n-k;i<n;i++) y[p++]=i;23         for(int i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;24         25         for(int i=0;i<m;i++) c[i]=0;26         for(int i=0;i<n;i++) c[x[y[i]]]++;27         for(int i=0;i<m;i++) c[i]+=c[i-1];28         for(int i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];29         30         swap(x,y);31         p=1;  x[sa[0]]=0;32         for(int i=1;i<n;i++)33              x[sa[i]]=y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;34         if(p>=n)  break;35         m=p;36     }37 }38 //构造height数组表示lcp(suffix(sa[i-1]),suffix(sa[i]))39 int rank[maxn],height[maxn];40 void getHeight() {41     int i,j,k=0;42     for(int i=0;i<n;i++) rank[sa[i]]=i;43     for(int i=0;i<n;i++) {44         if(k) k--;45         int j=sa[rank[i]-1];46         while(s[i+k]==s[j+k]) k++;47         height[rank[i]]=k;48     }49 }50 51 bool can(int k) {52     int min=sa[0],max=sa[0];53     for(int i=1;i<n;i++) {54         if(height[i]<k) min=max=sa[i];  //新组的开始55         if(sa[i]<min) min=sa[i];56         if(sa[i]>max) max=sa[i];57         if(max-min>=k) return true;58     }59     return false;60 }61 62 int main() {63     while(scanf("%d",&n)==1 && n) 64     {65         for(int i=0;i<n;i++)scanf("%d",&s[i]);66         for(int i=n-1;i>0;i--)s[i]=s[i]-s[i-1]+100;67         n--;//减少一个长度68         for(int i=0;i<n;i++)s[i]=s[i+1];69         s[n]=0;        //n==1时防止 RE 70         build_sa(200);71         getHeight();72         int L=0,R=n/2;73         while(L<R) {74             int M=L+(R-L+1)/2;75             if(can(M)) L=M;  else R=M-1;76         }77         L++;                        //对应有L+1个音符 78         if(L<=4) printf("0\n");79         else printf("%d\n",L);80     }81     return 0;82 }

2016-2-19

【思路】

SAM+DP

处理出right集的最大值mx和最小值mn，由p更新fa，则该重复子串的长度为min{l[i],mx[i]-mn[i]}

【代码】

 1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 using namespace std; 5  6 const int N = 4*1e4+10; 7 const int sigma = 180; 8  9 int s[N/2];10 int root,last,sz,ch[N][sigma],fa[N],l[N],mn[N],mx[N];11 int b[N],cnt[N],n;12 13 void init() {14     sz=0; root=last=++sz;15     memset(fa,0,sizeof(fa));16     memset(mx,0,sizeof(mx));17     memset(mn,127,sizeof(mn));18     memset(cnt,0,sizeof(cnt));19     memset(ch,0,sizeof(ch));20 }21 void add(int x) {22     int c=s[x];23     int p=last,np=++sz; last=np;24     mn[np]=mx[np]=l[np]=x;25     for(;p&&!ch[p][c];p=fa[p]) ch[p][c]=np;26     if(!p) fa[np]=root;27     else {28         int q=ch[p][c];29         if(l[p]+1==l[q]) fa[np]=q;30         else {31             int nq=++sz; l[nq]=l[p]+1;32             memcpy(ch[nq],ch[q],sizeof(ch[q]));33             fa[nq]=fa[q];34             fa[np]=fa[q]=nq;35             for(;p&&q==ch[p][c];p=fa[p]) ch[p][c]=nq;36         }37     }38 }39 void solve() {40     for(int i=1;i<=sz;i++) cnt[l[i]]++;41     for(int i=1;i<=n;i++) cnt[i]+=cnt[i-1];42     for(int i=1;i<=sz;i++) b[cnt[l[i]]--]=i;43     int ans=0;44     for(int i=sz;i;i--) {45         int p=b[i];46         if(fa[p]) {47             if(mn[fa[p]]>mn[p]) mn[fa[p]]=mn[p];48             if(mx[fa[p]]<mx[p]) mx[fa[p]]=mx[p];49         }50     }51     for(int i=1;i<=sz;i++)52         ans=max(ans,min(l[i],mx[i]-mn[i]));53     if(ans<4) puts("0");54     else printf("%d\n",ans+1);55 }56 void read(int& x) {57     char c=getchar(); int f=1; x=0;58     while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}59     while(isdigit(c)) x=x*10+c-'0',c=getchar();60     x*=f;61 }62 63 int main() {64     while(read(n),n) {65         init();66         for(int i=1;i<=n;i++) read(s[i]); n--;67         for(int i=1;i<=n;i++) s[i]=s[i+1]-s[i]+88,add(i);68         solve();69     }70     return 0;71 }

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