后缀自动机三·重复旋律6

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\n", ans[i]);
120     return 0;
121 }

posted on 2020-09-22 20:24 唔哩Wulili 阅读(124) 评论(0) 编辑收藏举报