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- // Routine for computing the length of the longest common prefix of any two
- // suffixes
- const int maxn = 1000;
- struct SuffixArray {
- char s[maxn]; //original string, the last character must be 0 and there is no 0 before.
- int sa[maxn]; //sa[i] = the index of the i-th smallest suffix.
- int rrank[maxn]; //rank[i] = rank of the suffix of index i. rank[0] must be n-1.
- int height[maxn]; // height[i] = The length of the longest common prefix of sa[i-1] and sa[i]
- int t[maxn], t2[maxn], c[maxn]; // aux
- int n; // the number of character.
- void clear() {
- n = 0;
- memset(sa, 0, sizeof(sa));
- memset(t, 0 , sizeof(t));
- memset(t2, 0 , sizeof(t2));
- }
- // m = max(char) + 1;
- //Build the suffix array of string s, every character should be between 0 and m-1.
- // initialiser s and n before calling this function.
- void build_sa(int m) {
- int i, *x = t, *y = t2;
- for(i = 0; i < m; i++) c[i] = 0;
- for(i = 0; i < n; i++) c[x[i] = s[i]]++;
- for(i = 1; i < m; i++) c[i] += c[i-1];
- for(i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;
- for(int k = 1; k <= n; k <<= 1) {
- int p = 0;
- for(i = n-k; i < n; i++) y[p++] = i;
- for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i]-k;
- for(i = 0; i < m; i++) c[i] = 0;
- for(i = 0; i < n; i++) c[x[y[i]]]++;
- for(i = 0; i < m; i++) c[i] += c[i-1];
- for(i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
- swap(x, y);
- p = 1; x[sa[0]] = 0;
- for(i = 1; i < n; i++)
- x[sa[i]] = y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+k]==y[sa[i]+k] ? p-1 : p++;
- if(p >= n) break;
- m = p;
- }
- }
- void build_height() {
- int i, k = 0;
- for(i = 0; i < n; i++) rrank[sa[i]] = i;
- for(i = 0; i < n; i++) {
- if(k) k--;
- int j = sa[rrank[i]-1];
- while(s[i+k] == s[j+k]) k++;
- height[rrank[i]] = k;
- }
- }
- //The LCP of any two suffx = RMQ(height[i+1],...,height[j])
- };
- void test() {
- SuffixArray SA;
- strcpy(SA.s, "1122330");
- SA.clear();
- SA.n = strlen(SA.s);
- SA.build_sa(100);
- SA.build_height();
- for (int i = 0; i < SA.n; i++)
- cout << i << " " << SA.s[i] << " " << SA.sa[i] << " " << SA.rrank[i] << endl;
- }
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