对于一个长度为N的字符串，我们想知道每个长度的串的最长出现次数，所以这里就用到了后缀自动机了，知道后缀自动机中每个点表示的长度为len[link]+1~len，所以其实我们可以直接给ans[len]去取最大值，然后跑一个后缀最大值即可。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <string> 5 #include <cstring> 6 #include <algorithm> 7 #include <limits> 8 #include <vector> 9 #include <stack> 10 #include <queue> 11 #include <set> 12 #include <map> 13 #include <bitset> 14 #include <unordered_map> 15 #include <unordered_set> 16 #define lowbit(x) ( x&(-x) ) 17 #define pi 3.141592653589793 18 #define e 2.718281828459045 19 #define INF 0x3f3f3f3f 20 #define HalF (l + r)>>1 21 #define lsn rt<<1 22 #define rsn rt<<1|1 23 #define Lson lsn, l, mid 24 #define Rson rsn, mid+1, r 25 #define QL Lson, ql, qr 26 #define QR Rson, ql, qr 27 #define myself rt, l, r 28 #define pii pair<int, int> 29 #define MP(a, b) make_pair(a, b) 30 using namespace std; 31 typedef unsigned long long ull; 32 typedef unsigned int uit; 33 typedef long long ll; 34 const int maxN = 1e6 + 7; 35 int ans[maxN] = {0}; 36 struct SAM 37 { 38 struct state 39 { 40 int len, link, next[26]; 41 } st[maxN << 1]; 42 int siz = 1, last, dp[maxN << 1]; 43 void init() 44 { 45 siz = 1; 46 st[1].len = 0; 47 st[1].link = 0; 48 siz++; 49 last = 1; 50 } 51 int extend(int c) 52 { 53 int cur = siz++; 54 dp[cur] = 1; 55 st[cur].len = st[last].len + 1; 56 int p = last; 57 while (p != 0 && !st[p].next[c]) 58 { 59 st[p].next[c] = cur; 60 p = st[p].link; 61 } 62 if (p == 0) 63 { 64 st[cur].link = 1; 65 } 66 else 67 { 68 int q = st[p].next[c]; 69 if (st[p].len + 1 == st[q].len) 70 { 71 st[cur].link = q; 72 } 73 else 74 { 75 int clone = siz++; dp[clone] = 0; 76 st[clone].len = st[p].len + 1; 77 memcpy(st[clone].next, st[q].next, sizeof(st[q].next)); 78 st[clone].link = st[q].link; 79 while (p != 0 && st[p].next[c] == q) 80 { 81 st[p].next[c] = clone; 82 p = st[p].link; 83 } 84 st[q].link = st[cur].link = clone; 85 } 86 } 87 return last = cur; 88 } 89 vector<int> to[maxN << 1]; 90 void dfs(int u) 91 { 92 for(int v : to[u]) 93 { 94 dfs(v); 95 dp[u] += dp[v]; 96 } 97 ans[st[u].len] = max(ans[st[u].len], dp[u]); 98 } 99 void slove() 100 { 101 for(int i=0; i<siz; i++) to[i].clear(); 102 for(int i=2; i<siz; i++) to[st[i].link].push_back(i); 103 dfs(1); 104 } 105 } sam; 106 char s[maxN]; 107 int main() 108 { 109 scanf("%s", s); 110 sam.init(); 111 int len = (int)strlen(s); 112 for(int i=0; i<len; i++) 113 { 114 sam.extend(s[i] - 'a'); 115 ans[i + 1] = 0; 116 } 117 sam.slove(); 118 for(int i=len - 1; i>=1; i--) ans[i] = max(ans[i], ans[i + 1]); 119 for(int i=1; i<=len; i++) printf("%d ", ans[i]); 120 return 0; 121 }