SPOJ NSUBSTR(Substrings-字尾自動機統計串出現次數-Right集合&Parent樹の暴走)
阿新 • • 發佈：2019-01-31
8222. SubstringsProblem code: NSUBSTR |
You are given a string S which consists of 250000 lowercase latin letters at most. We define F(x) as the maximal number of times that some string with length x appears in S. For example for string 'ababa' F(3) will be 2 because there is a string 'aba' that occurs twice. Your task is to output F(i) for every i so that 1<=i<=|S|.
Input
String S consists of at most 250000 lowercase latin letters.
Output
Output |S| lines. On the i-th line output F(i).
Example
Input: ababa Output: 3 2 2 1 1
字尾自動機入門用。。。
有關字尾自動機的話題省略100字、、（我能告訴你我其實也沒懂嗎）
樹邊的結點表示字尾所在Right集合的點=葉結點(定義)
故把每個結點的|Right|(集合大小)求出來更新答案
得到 結點代表的最長字尾=k的最大出現次數
→ 結點代表的最長字尾≥k的最大出現次數
#include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<functional> #include<iostream> #include<cmath> #include<cctype> #include<ctime> using namespace std; #define For(i,n) for(int i=1;i<=n;i++) #define Fork(i,k,n) for(int i=k;i<=n;i++) #define Rep(i,n) for(int i=0;i<n;i++) #define ForD(i,n) for(int i=n;i;i--) #define RepD(i,n) for(int i=n;i>=0;i--) #define Forp(x) for(int p=pre[x];p;p=next[p]) #define Lson (x<<1) #define Rson ((x<<1)+1) #define MEM(a) memset(a,0,sizeof(a)); #define MEMI(a) memset(a,127,sizeof(a)); #define MEMi(a) memset(a,128,sizeof(a)); #define INF (2139062143) #define F (100000007) #define MAXN (600000+10) long long mul(long long a,long long b){return (a*b)%F;} long long add(long long a,long long b){return (a+b)%F;} long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;} typedef long long ll; struct node { int pre,ch[26]; int step; char c; node(char c):c(c){pre=step=0;memset(ch,0,sizeof(ch)); } node():c(0){pre=step=0;memset(ch,0,sizeof(ch)); } }a[MAXN]; int last=0,total=0; void ins(char c) { int np=++total;a[np].c=c,a[np].step=a[last].step+1; int p=last; for(;!a[p].ch[c];p=a[p].pre) a[p].ch[c]=np; if (a[p].ch[c]==np) a[np].pre=p; else { int q=a[p].ch[c]; if (a[q].step>a[p].step+1) { int nq=++total;a[nq]=a[q];a[nq].step=a[p].step+1; a[np].pre=a[q].pre=nq; for(;a[p].ch[c]==q;p=a[p].pre) a[p].ch[c]=nq; }else a[np].pre=q; } last=np; } char dfsc[MAXN]=""; void dfs(int x,int l) { dfsc[l]=a[x].c+'a'; dfsc[l+1]=0; // if (dfsc[1]) printf("%s\n",dfsc+1); // Rep(i,26) if (a[x].ch[i]) dfs(a[x].ch[i],l+1); } char s[MAXN]; int n,t[MAXN]={0},r[MAXN]={0},v[MAXN]={0},ans[MAXN]={0}; int main() { // freopen("spojNSUBSTR.in","r",stdin); scanf("%s",s+1);int n=strlen(s+1); For(i,n) ins(s[i]-'a'); // dfs(0,0); For(i,total) t[a[i].step]++; For(i,n) t[i]+=t[i-1]; For(i,total) r[t[a[i].step]--]=i; // For(i,total) cout<<r[i]<<' ';cout<<endl; int now=0; For(i,n) v[now=a[now].ch[s[i]-'a']]++; ForD(i,total) { int now=r[i]; ans[a[now].step]=max(ans[a[now].step],v[now]); v[a[now].pre]+=v[now]; } ForD(i,n-1) ans[i]=max(ans[i],ans[i+1]); For(i,n) printf("%d\n",ans[i]); return 0; }