NSUBSTR Substrings
阿新 • • 發佈:2018-12-15
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
題解:SAM模板題。
程式碼:
#include<cstdio> #include<vector> #include<cstring> #include<algorithm> using namespace std; typedef long long LL; const int MAXN = 1000005; typedef struct Node { int len, pre;//pre:對應圖片中的綠線(一個節點顯然最多隻有一條綠線) int Next[26];//Next[]:對應圖片中的藍線 }Node; Node tre[MAXN*2]; vector<int> G[MAXN*2];//用於建Suffix Links樹 int cnt, last, siz[MAXN*2], ans[MAXN*2]; char str[MAXN]; void Insert(char ch) { int p, q, now, rev; p = last, now = cnt++; tre[now].len = tre[last].len+1; siz[now]++; //如果節點u包含子串S[1..i],那麼滿足|siz(u)| = ∑|siz(son(u))|+1,這裡先加上那個1,這樣的話對於SL樹就可以直接求和了 while(p!=-1 && tre[p].Next[ch-'a']==0) //每次跳到當前最長且siz集合與當前不同的字尾上,對應圖片中的綠線回退(例如7→8→5→S) { tre[p].Next[ch-'a'] = now; //tran(st[p], ch)=now,對應圖片中的藍線連線 p = tre[p].pre; } if(p==-1) tre[now].pre = 0; //情況①,遞迴到了初始節點S(空子串)(例如圖片中的9號節點pre[9]=0) else //如果中途某個子串tran(st[p], ch)已經存在 { q = tre[p].Next[ch-'a']; if(tre[q].len==tre[p].len+1) //情況②:節點q的最長子串剛好就是節點p的最長子串+S[i],也就是len[q] = len[p]+1 tre[now].pre = q; else //情況③ { rev = cnt++; tre[rev] = tre[q]; tre[rev].len = tre[p].len+1; //這三行就是對Suffix Links內向樹的插點操作 tre[q].pre = tre[now].pre = rev; while(p!=-1 && tre[p].Next[ch-'a']==q) { tre[p].Next[ch-'a'] = rev; p = tre[p].pre; } } } last = now; } void SechSL(int u) //求出所有節點的|endpos()| { int v; for(int i=0 ; i<G[u].size() ; ++i) { v = G[u][i]; SechSL(v); siz[u] += siz[v]; } ans[tre[u].len] = max(ans[tre[u].len], siz[u]); } inline void Init() { cnt = last = 0; memset(tre, 0, sizeof(tre)); for(int i=0 ; i<MAXN*2 ; ++i)G[i].clear(); tre[cnt++].pre = -1; } int main() { int n; scanf("%s", str+1); n = strlen(str+1); Init(); for(int i=1 ; i<=n ; ++i)Insert(str[i]); for(int i=1 ; i<=cnt-1 ; ++i)G[tre[i].pre].push_back(i);//建樹 SechSL(0); /*--------------------------- LL ans = 0; for(i=1;i<=cnt-1;i++) ans += tre[i].len-tre[tre[i].pre].len; //求出有多少個本質不同的子串 -----------------------------*/ for(int i=n ; i>=1 ; i--) ans[i] = max(ans[i], ans[i+1]); for(int i=1 ; i<=n ; ++i) printf("%d\n", ans[i]); return 0; }