Cyclical Quest CodeForces - 235C 后缀自动机

题意：

给出一个字符串，给出一些子串，问每个子串分别在母串中圆环匹配的次数，

圆环匹配的意思是将该子串拆成两段再首位交换相接的串和母串匹配，比

如aaab变成baaa,abaa,aaba再进行匹配。

题解：

如何求出所有的循环串出现的次数呢？

先将S串放入后缀自动机

把查询串扩大一倍，然后在后缀自动机上去匹配，

只要匹配长度大于子串长度小于2倍子串长度的，必然对应这一种循环串

记录一下

// 每个节点子串出现的次数
 
统计答案即可

 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <unordered_map>
 
 #define  pi    acos(-1.0)
 #define  eps   1e-9
 #define  fi    first
 #define  se    second
 #define  rtl   rt<<1
 #define  rtr   rt<<1|1
 #define  bug                printf("******\n")
 #define  mem(a, b)          memset(a,b,sizeof(a))
 #define  name2str(x)        #x
 #define  fuck(x)            cout<<#x" = "<<x<<endl
 #define  sfi(a)             scanf("%d", &a)
 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  sfL(a)             scanf("%lld", &a)
 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 #define  sfs(a)             scanf("%s", a)
 #define  sffs(a, b)         scanf("%s %s", a, b)
 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 #define  FIN                freopen("../in.txt","r",stdin)
 #define  gcd(a, b)          __gcd(a,b)
 #define  lowbit(x)          x&-x
 #define  IO                 iOS::sync_with_stdio(false)
 
 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 const ULL seed = ;
 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = 1e9 + ;
 const int maxn = ;
 char s[maxn];
 int Q;
 
 struct Suffix_Automaton {
     int last, tot, nxt[maxn << ][], fail[maxn << ];//last是未加入此字符前最长的前缀（整个串）所属的节点的编号
     int len[maxn << ];// 最长子串的长度 (该节点子串数量 = len[x] - len[fa[x]])
     int sa[maxn << ], c[maxn << ];
     int sz[maxn << ];// 被后缀链接的个数，方便求节点字符串的个数
     LL num[maxn << ];// 该状态子串的数量
     LL maxx[maxn << ];// 长度为x的子串出现次数最多的子串的数目
     LL sum[maxn << ];// 该节点后面所形成的自字符串的总数
     LL subnum, sublen;// subnum表示不同字符串数目,sublen表示不同字符串总长度
     int X[maxn << ], Y[maxn << ]; // Y表示排名为x的节点，X表示该长度前面还有多少个
     int minn[maxn << ], mx[maxn << ];//minn[i]表示多个串在后缀自动机i节点最长公共子串，mx[i]表示单个串的最长公共子串
     void init() {
         tot = last = ;
         fail[] = len[] = ;
         for (int i = ; i < ; i++) nxt[][i] = ;
     }
 
     void extend(int c) {
         int u = ++tot, v = last;
         len[u] = len[v] + ;
         num[u] = ;
         for (; v && !nxt[v][c]; v = fail[v]) nxt[v][c] = u;
         if (!v) fail[u] = , sz[]++;
         else if (len[nxt[v][c]] == len[v] + ) fail[u] = nxt[v][c], sz[nxt[v][c]]++;
         else {
             int now = ++tot, cur = nxt[v][c];
             len[now] = len[v] + ;
             memcpy(nxt[now], nxt[cur], sizeof(nxt[cur]));
             fail[now] = fail[cur];
             fail[cur] = fail[u] = now;
             for (; v && nxt[v][c] == cur; v = fail[v]) nxt[v][c] = now;
             sz[now] += ;
         }
         last = u;
         //return len[last] - len[fail[last]];//多添加一个子串所产生不同子串的个数
     }
 
     void get_num() {// 每个节点子串出现的次数
         for (int i = ; i <= tot; i++) X[len[i]]++;
         for (int i = ; i <= tot; i++) X[i] += X[i - ];
         for (int i = ; i <= tot; i++) Y[X[len[i]]--] = i;
         for (int i = tot; i >= ; i--) num[fail[Y[i]]] += num[Y[i]];
     }
 
     void get_maxx(int n) {// 长度为x的子串出现次数最多的子串的数目
         get_num();
         for (int i = ; i <= tot; i++) maxx[len[i]] = max(maxx[len[i]], num[i]);
     }
 
     void get_sum() {// 该节点后面所形成的自字符串的总数
         get_num();
         for (int i = tot; i >= ; i--) {
             sum[Y[i]] = ;
             for (int j = ; j <= ; j++)
                 sum[Y[i]] += sum[nxt[Y[i]][j]];
         }
     }
 
     void get_subnum() {//本质不同的子串的个数
         subnum = ;
         for (int i = ; i <= tot; i++) subnum += len[i] - len[fail[i]];
     }
 
     void get_sublen() {//本质不同的子串的总长度
         sublen = ;
         for (int i = ; i <= tot; i++) sublen += 1LL * (len[i] + len[fail[i]] + ) * (len[i] - len[fail[i]]) / ;
     }
 
     void get_sa() { //获取sa数组
         for (int i = ; i <= tot; i++) c[len[i]]++;
         for (int i = ; i <= tot; i++) c[i] += c[i - ];
         for (int i = tot; i >= ; i--) sa[c[len[i]]--] = i;
     }
 
     void match(char s[]) {//多个串的最长公共子串
         mem(mx, );
         int n = strlen(s), p = , maxlen = ;
         for (int i = ; i < n; i++) {
             int c = s[i] - 'a';
             if (nxt[p][c]) p = nxt[p][c], maxlen++;
             else {
                 for (; p && !nxt[p][c]; p = fail[p]);
                 if (!p) p = , maxlen = ;
                 else maxlen = len[p] + , p = nxt[p][c];
             }
             mx[p] = max(mx[p], maxlen);
         }
         for (int i = tot; i; i--)
             mx[fail[i]] = max(mx[fail[i]], min(len[fail[i]], mx[i]));
         for (int i = tot; i; i--)
             if (minn[i] == - || minn[i] > maxx[i]) minn[i] = mx[i];
     }
 
     void get_kth(int k) {//求出字典序第K的子串
         int pos = , cnt;
         string s = "";
         while (k) {
             for (int i = ; i <= ; i++) {
                 if (nxt[pos][i] && k) {
                     cnt = nxt[pos][i];
                     if (sum[cnt] < k) k -= sum[cnt];
                     else {
                         k--;
                         pos = cnt;
                         s += (char) (i + 'a');
                         break;
                     }
                 }
             }
         }
         cout << s << endl;
     }
 
     int vis[maxn << ];
 
     void solve(int lenlen, int id) {
         int maxlen = , p = , ans = ;
         for (int i = ; i <= lenlen * ; i++) {
             int c = s[i] - 'a';
             if (nxt[p][c]) maxlen++, p = nxt[p][c];
             else {
                 for (; p && !nxt[p][c]; p = fail[p]);
                 if (p == ) p = , maxlen = ;
                 else maxlen = len[p] + , p = nxt[p][c];
             }
             if (i >= lenlen && maxlen >= lenlen) {
                 int temp = p;
                 while (temp && !(len[fail[temp]] +  <= lenlen && len[temp] >= lenlen))temp = fail[temp];
                 if (!temp) temp = ;
                 if (vis[temp] != id) vis[temp] = id, ans += num[temp];
             }
         }
         printf("%d\n", ans);
     }
 } sam;
 
 int main() {
 #ifndef ONLINE_JUDGE
     FIN;
 #endif
     sam.init();
     sfs(s + );
     int n = strlen(s + );
     for (int i = ; i <= n; i++) sam.extend((s[i] - 'a'));
     sam.get_num();
     sfi(Q);
     for (int i = ; i <= Q; i++) {
         sfs(s + );
         int len = strlen(s + );
         for (int i = ; i <= len; i++) s[i + len] = s[i];
         s[ * len + ] = '\0';
         //fuck(s + 1);
         sam.solve(len, i);
     }
 #ifndef ONLINE_JUDGE
     cout << "Totle Time : " << (double) clock() / CLOCKS_PER_SEC << "s" << endl;
 #endif
     return ;
 }