1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <vector>
6 using namespace std;
7 const int maxn = 200;
8 char s[maxn];
9 int sa[maxn], t[maxn], t2[maxn], c[maxn];
10 int n;
11 //构造字符串s的后缀数组， 每个字符值必须为0 ～ m-1
12 void build_sa(int m) {
13 int *x = t, *y = t2;
14 //基数排序
15 for(int i = 0; i < m; i++) c[i] = 0;
16 for(int i = 0; i < n; i++) c[x[i] = s[i]]++;
17 for(int i = 1; i < m; i++) c[i] += c[i-1];
18 for(int i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;
19 for(int k = 1; k <= n; k <<= 1) {
20 int p = 0;
21 //直接利用sa数组排序第二关键字
22 for(int i = n-k; i < n; i++) y[p++] = i;
23 for(int i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
24 //基数排序第一关键字
25 for(int i = 0; i < m; i++) c[i] = 0;
26 for(int i = 0; i < n; i++) c[x[y[i]]]++;
27 for(int i = 1; i < m; i++) c[i] += c[i-1];
28 for(int i = n-1; i>= 0; i--) sa[--c[x[y[i]]]] = y[i];
29 //根据sa和y数组计算新的x数组
30 swap(x, y);
31 p = 1;
32 x[sa[0]] = 0;
33 for(int i = 1; i < n; i++)
34 x[sa[i]] = (y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++);
35 if(p >= n) break;
36 m = p;
37 }
38 }
39
40 int rank_[maxn]; //rank[i]代表后缀i在sa数组中的下标
41 int height[maxn]; //height[i] 定义为sa[i-1] 和 sa[i] 的最长公共前缀
42 //后缀j和k的LCP长度等于RMQ(height, rank[j]+1, rank[k])
43 void get_height() {
44 int i, j, k = 0;
45 for(int i = 0; i < n; i++) rank_[sa[i]] = i;
46 for(int i = 0; i < n; i++) {
47 if(!rank_[i]) continue;
48 int j = sa[rank_[i]-1];
49 if(k) k--;
50
51 while(s[i+k] == s[j+k]) k++;
52 height[rank_[i]] = k;
53 }
54 }
55 int d[maxn][50];
56 void rmq_init() {
57 for(int i = 0; i < n; i++) d[i][0] = height[i];
58 for(int j = 1; (1<<j) <= n; j++)
59 for(int i = 0; i + (1<<j) - 1 < n; i++)
60 d[i][j] = min(d[i][j-1], d[i+(1<<(j-1))][j-1]);
61 }
62 int rmq(int l, int r) {
63 if(l == r) return n-l;
64 if(rank_[l] > rank_[r]) swap(l, r);
65 int L = rank_[l]+1;
66 int R = rank_[r];
67 int k = 0;
68 while((1<<(k+1)) <= R-L+1) k++;
69 return min(d[L][k], d[R-(1<<k)+1][k]);
70 }
71 //LCP加速多模式匹配
72 int m;
73 int cmp_suffix(char* P, int p, int c,int &k) {
74 k = 0;
75 int i;
76 for(i = 0; P[c+i] == s[sa[p]+c+i]; i++) {
77 if(P[c+i] == '\0')
78 return 0;
79 k++;
80 }
81 if(P[c+i] == '\0')
82 return 0;
83 return P[c+i] - s[sa[p]+c+i];
84 }
85 vector<int> A;
86 void b_search(char*P, int L, int R) {
87 int k;
88 if(cmp_suffix(P, L, 0, k) < 0) return ;
89 if(cmp_suffix(P, R, 0, k) > 0) return ;
90 int c = 0, rr = 0;
91 int lst = -1;
92 k = 0;
93 while(R >= L) {
94 int M = L + (R-L)/2;
95 if(lst != -1) c = rmq(lst, sa[M]);
96 if(c <= k) {
97 int res = cmp_suffix(P, M, c, k);
98 rr = res;
99 if(!res) {
100 A.push_back(sa[M]);
101 b_search(P, L, M-1);
102 b_search(P, M+1, R);
103 return;
104 }
105 lst = sa[M];
106 if(res < 0) R = M-1; else L = M+1;
107 }
108 else if(rr < 0)R = M-1;
109 else L = M+1;
110 }
111 }
112 void find(char* P) { //找到全部的匹配位置存入A数组中
113 A.clear();
114 m = strlen(P);
115 int L = 0, R = n-1;
116 b_search(P, L, R);
117 sort(A.begin(), A.end());
118 }
