Boring counting HDU - 3518 后缀自动机

题意：

对于给出的字符串S, 长度不超过1000,

求其中本质不同的子串的数量, 这些子串满足在字符串S中出现了至少不重合的2次

题解：

将串放入后缀自动机中然后求出每一个节点对应的子串为后缀的子串出现的最早和最晚的位置

然后根据

//return len[last] - len[fail[last]];//多添加一个子串所产生不同子串的个数

然后根据这个改一下  

if (R[i] - L[i] > len[fail[i]]) ans += min(len[i], R[i] - L[i]) - len[fail[i]];

 #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 maxn = 1e4 + ;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = 1e9 + ;

 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]])
     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 L[maxn << ], R[maxn << ];//L表示对应节点代表的数组的最早出现位置,R表示最晚出现位置

     void init() {
         tot = last = ;
         fail[] = len[] = ;
         for (int i = ; i <= ; i++) nxt[][i] = ;

     }

     void extend(int c) {
         int u = ++tot, v = last;
         for (int i = ; i <= ; i++) nxt[u][i] = ;
         fail[u] = ;
         L[u] = R[u] = len[u] = len[v] + ;
         num[u] = ;
         for (; v && !nxt[v][c]; v = fail[v]) nxt[v][c] = u;
         if (!v) fail[u] = ;
         else if (len[nxt[v][c]] == len[v] + ) fail[u] = 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;
             L[now] = L[cur], R[now] = R[cur];
             for (; v && nxt[v][c] == cur; v = fail[v]) nxt[v][c] = now;
         }
         last = u;
         //return len[last] - len[fail[last]];
     }

     void get_sa() { // Y表示排名为x的节点，X表示该长度前面还有多少个
         for (int i = ; i <= tot; i++) X[i] = ;
         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;
     }

     int get_L_R() {
         int ans = ;
         for (int i = tot; i; i--) {
             L[fail[Y[i]]] = min(L[fail[Y[i]]], L[Y[i]]);
             R[fail[Y[i]]] = max(R[fail[Y[i]]], R[Y[i]]);
         }
         for (int i = ; i <= tot; i++)//不相交且出现次数至少2的子串个数
             if (R[i] - L[i] > len[fail[i]]) ans += min(len[i], R[i] - L[i]) - len[fail[i]];
         return ans;
     }
 } sam;

 char s[maxn];

 int main() {
 #ifndef ONLINE_JUDGE
     FIN;
 #endif
     while (~sfs(s)) {
         if (s[] == '#') break;
         sam.init();
         int len = strlen(s);
         for (int i = ; i < len; ++i) sam.extend((s[i] - 'a'));
         sam.get_sa();
         printf("%d\n", sam.get_L_R());
     }
 #ifndef ONLINE_JUDGE
     cout << "Totle Time : " << (double) clock() / CLOCKS_PER_SEC << "s" << endl;
 #endif
     return ;
 }