啥都不会啊!怎么办啊!

Fitz

慢慢来生活总会好起来的!!!

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]];

  1 #include <set>
  2 #include <map>
  3 #include <stack>
  4 #include <queue>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <cstdio>
  8 #include <string>
  9 #include <vector>
 10 #include <cstring>
 11 #include <iostream>
 12 #include <algorithm>
 13 #include <unordered_map>
 14 
 15 #define  pi    acos(-1.0)
 16 #define  eps   1e-9
 17 #define  fi    first
 18 #define  se    second
 19 #define  rtl   rt<<1
 20 #define  rtr   rt<<1|1
 21 #define  bug                printf("******\n")
 22 #define  mem(a, b)          memset(a,b,sizeof(a))
 23 #define  name2str(x)        #x
 24 #define  fuck(x)            cout<<#x" = "<<x<<endl
 25 #define  sfi(a)             scanf("%d", &a)
 26 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 27 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 28 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 29 #define  sfL(a)             scanf("%lld", &a)
 30 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 31 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 32 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 33 #define  sfs(a)             scanf("%s", a)
 34 #define  sffs(a, b)         scanf("%s %s", a, b)
 35 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 36 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 37 #define  FIN                freopen("../in.txt","r",stdin)
 38 #define  gcd(a, b)          __gcd(a,b)
 39 #define  lowbit(x)          x&-x
 40 #define  IO                 iOS::sync_with_stdio(false)
 41 
 42 
 43 using namespace std;
 44 typedef long long LL;
 45 typedef unsigned long long ULL;
 46 const ULL seed = 13331;
 47 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 48 const int maxn = 1e4 + 50;
 49 const int maxm = 8e6 + 10;
 50 const int INF = 0x3f3f3f3f;
 51 const int mod = 1e9 + 7;
 52 
 53 struct Suffix_Automaton {
 54     int last, tot, nxt[maxn << 1][26], fail[maxn << 1];//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
 55     int len[maxn << 1];// 最长子串的长度 (该节点子串数量 = len[x] - len[fa[x]])
 56     LL num[maxn << 1];// 该状态子串的数量
 57     LL maxx[maxn << 1];// 长度为x的子串出现次数最多的子串的数目
 58     LL sum[maxn << 1];// 该节点后面所形成的自字符串的总数
 59     LL subnum, sublen;// subnum表示不同字符串数目,sublen表示不同字符串总长度
 60     int X[maxn << 1], Y[maxn << 1]; // Y表示排名为x的节点,X表示该长度前面还有多少个
 61     int L[maxn << 1], R[maxn << 1];//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
 62 
 63     void init() {
 64         tot = last = 1;
 65         fail[1] = len[1] = 0;
 66         for (int i = 0; i <= 25; i++) nxt[1][i] = 0;
 67 
 68     }
 69 
 70     void extend(int c) {
 71         int u = ++tot, v = last;
 72         for (int i = 0; i <= 25; i++) nxt[u][i] = 0;
 73         fail[u] = 0;
 74         L[u] = R[u] = len[u] = len[v] + 1;
 75         num[u] = 1;
 76         for (; v && !nxt[v][c]; v = fail[v]) nxt[v][c] = u;
 77         if (!v) fail[u] = 1;
 78         else if (len[nxt[v][c]] == len[v] + 1) fail[u] = nxt[v][c];
 79         else {
 80             int now = ++tot, cur = nxt[v][c];
 81             len[now] = len[v] + 1;
 82             memcpy(nxt[now], nxt[cur], sizeof(nxt[cur]));
 83             fail[now] = fail[cur];
 84             fail[cur] = fail[u] = now;
 85             L[now] = L[cur], R[now] = R[cur];
 86             for (; v && nxt[v][c] == cur; v = fail[v]) nxt[v][c] = now;
 87         }
 88         last = u;
 89         //return len[last] - len[fail[last]];
 90     }
 91 
 92 
 93     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
 94         for (int i = 0; i <= tot; i++) X[i] = 0;
 95         for (int i = 1; i <= tot; i++) X[len[i]]++;
 96         for (int i = 1; i <= tot; i++) X[i] += X[i - 1];
 97         for (int i = 1; i <= tot; i++) Y[X[len[i]]--] = i;
 98     }
 99 
100     int get_L_R() {
101         int ans = 0;
102         for (int i = tot; i; i--) {
103             L[fail[Y[i]]] = min(L[fail[Y[i]]], L[Y[i]]);
104             R[fail[Y[i]]] = max(R[fail[Y[i]]], R[Y[i]]);
105         }
106         for (int i = 1; i <= tot; i++)//不相交且出现次数至少2的子串个数
107             if (R[i] - L[i] > len[fail[i]]) ans += min(len[i], R[i] - L[i]) - len[fail[i]];
108         return ans;
109     }
110 } sam;
111 
112 char s[maxn];
113 
114 int main() {
115 #ifndef ONLINE_JUDGE
116     FIN;
117 #endif
118     while (~sfs(s)) {
119         if (s[0] == '#') break;
120         sam.init();
121         int len = strlen(s);
122         for (int i = 0; i < len; ++i) sam.extend((s[i] - 'a'));
123         sam.get_sa();
124         printf("%d\n", sam.get_L_R());
125     }
126 #ifndef ONLINE_JUDGE
127     cout << "Totle Time : " << (double) clock() / CLOCKS_PER_SEC << "s" << endl;
128 #endif
129     return 0;
130 }
View Code

 

posted @ 2019-09-23 19:59  Fitz~  阅读(219)  评论(0编辑  收藏  举报