POJ1743 Musical Theme （后缀数组 & 后缀自动机）最大不重叠相似子串
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.
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
Hint
Use scanf instead of cin to reduce the read time.
题解：最大不重叠相似子串。
先相邻元素做差，得到字符串。
后缀数组：利用后缀数组的height数组的性质，字典序相邻的字符串的相似度是最大的。
然后二分答案len，找相邻的height数组大于等于len的最大长度，如果最大长度>=k-1；
则是满足题意的答案，否则输出0；
后缀自动机：后缀自动机是利用endpos等价类的性质，对于每一个等价类，我们分别求出该类的
最大长度所在的位置和最小长度所在的位置，然后如果这两个位置的差不小于longest[i],则满足题意，
在这些满足题意的值中去最大值即可。
参考代码：
SA（后缀数组）
#include<iostream> #include<cstdio> #include<cstring> #include<string> using namespace std; typedef long long ll; const int INF=0x3f3f3f3f; const int maxn=20020; int n,s[maxn]; int rk[maxn],sa[maxn],height[maxn]; int x[maxn<<1],y[maxn<<1],c[maxn]; inline void get_SA(int m) { for(int i=1;i<=m;++i) c[i]=0; for(int i=1;i<=n;++i) ++c[x[i]=s[i]]; for(int i=2;i<=m;++i) c[i]+=c[i-1]; for(int i=n;i>=1;--i) sa[c[x[i]]--]=i; for(int k=1;k<=n;k<<=1) { int num=0; for(int i=n-k+1;i<=n;++i) y[++num]=i; for(int i=1;i<=n;++i) if(sa[i]>k) y[++num]=sa[i]-k; for(int i=1;i<=m;++i) c[i]=0; for(int i=1;i<=n;++i) ++c[x[i]]; for(int i=2;i<=m;++i) c[i]+=c[i-1]; for(int i=n;i>=1;--i) sa[c[x[y[i]]]--]=y[i],y[i]=0; swap(x,y); x[sa[1]]=1; num=1; for(int i=2;i<=n;++i) x[sa[i]]=(y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k])?num:++num; if(num==n) break; m=num; } } inline void get_height() { int k=0; for(int i=1;i<=n;++i) rk[sa[i]]=i; for(int i=1;i<=n;++i) { if(rk[i]==1) continue; if(k) --k; int j=sa[rk[i]-1]; while(j+k<=n&&i+k<=n&&s[i+k]==s[j+k]) ++k; height[rk[i]]=k; } } bool check(int k) { int mx=-INF,mi=INF; for(int i=1;i<=n;++i) { if(height[i]>=k) { mx=max(mx,max(sa[i],sa[i-1])); mi=min(mi,min(sa[i],sa[i-1])); if(mx-mi>k) return true; } else mx=-INF,mi=INF; } return false; } int main() { while(scanf("%d",&n) && n) { int pre,now; n--; scanf("%d",&pre); for(int i=1;i<=n;++i) scanf("%d",&now),s[i]=now-pre+88,pre=now; get_SA(176); get_height(); int l=1,r=n>>1; while(l+1<r) { int mid=l+r>>1; if(check(mid)) l=mid; else r=mid; } int ans; if(check(r)) ans=r; else ans=l; printf("%d\n",ans>=4? ans+1:0); } return 0; }
后缀自动机
/********* 后缀自动机做法 ***********/ #include<map> #include<ctime> #include<queue> #include<cmath> #include<cstdio> #include<vector> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define inf 1000000000 #define mod 1000000007 #define pa pair<int,int> #define ll long long using namespace std; int 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 n,a[20005]; struct sam{ int last,cnt,ans; int l[40005],r[40005],mx[40005],fa[40005],a[40005][180]; int q[40005],v[40005]; void init() { memset(l,127,sizeof(l)); memset(r,0,sizeof(r)); memset(v,0,sizeof(v)); memset(mx,0,sizeof(mx)); memset(fa,0,sizeof(fa)); memset(a,0,sizeof(a)); last=cnt=1;ans=0; } void extend(int c) { int p=last,np=last=++cnt;mx[np]=mx[p]+1; l[np]=r[np]=mx[np]; while(!a[p][c]&&p)a[p][c]=np,p=fa[p]; if(!p)fa[np]=1; else { int q=a[p][c]; if(mx[p]+1==mx[q]) fa[np]=q; else { int nq=++cnt;mx[nq]=mx[p]+1; memcpy(a[nq],a[q],sizeof(a[q])); fa[nq]=fa[q]; fa[np]=fa[q]=nq; while(a[p][c]==q) a[p][c]=nq,p=fa[p]; } } } void solve() { for(int i=1;i<=cnt;i++) v[mx[i]]++; for(int i=1;i<=n;i++) v[i]+=v[i-1]; for(int i=cnt;i;i--) q[v[mx[i]]--]=i; for(int i=cnt;i;i--) { int p=q[i]; l[fa[p]]=min(l[fa[p]],l[p]); r[fa[p]]=max(r[fa[p]],r[p]); } for(int i=1;i<=cnt;i++) ans=max(ans,min(mx[i],r[i]-l[i])); if(ans<4)puts("0"); else printf("%d\n",ans+1); } } sam; int main() { while(scanf("%d",&n)) { if(n==0)break; for(int i=1;i<=n;i++) a[i]=read();n--; for(int i=1;i<=n;i++) a[i]=a[i+1]-a[i]+88; sam.init(); for(int i=1;i<=n;i++) sam.extend(a[i]); sam.solve(); } return 0; }
