spoj 694 Distinct Substrings (后缀数组)

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spoj 694 Distinct Substrings

题意：
给出S, 求不重复子串的个数

限制：
|S| <= 1000

思路：
利用lcp进行统计

/*spoj 694 Distinct Substrings  题意：  给出S, 求不重复子串的个数  限制：  |S| <= 1000  思路：  利用lcp进行统计 */#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int N = 1005;int n;int times;int sa[N];int rank[N], tmp_rank[N];bool cmp_sa(int i, int j) {if(rank[i] != rank[j])return rank[i] < rank[j];else {int ri = i + times <= n ? rank[i + times] : -1;int rj = j + times <= n ? rank[j + times] : -1;return ri < rj;}}void build_sa(char *S, int *sa) {n = strlen(S);for(int i = 0; i <= n; ++i) {sa[i] = i;rank[i] = S[i];}for(times = 1; times <= n; times <<= 1) {sort(sa, sa + n + 1, cmp_sa);tmp_rank[sa[0]] = 0;for(int i = 1; i <= n; ++i) {tmp_rank[sa[i]] = tmp_rank[sa[i - 1]] + cmp_sa(sa[i - 1], sa[i]);}for(int i = 0; i <= n; ++i) {rank[i] = tmp_rank[i];}}}int lcp[N];void build_lcp(char *S, int *sa, int *lcp) {int h = 0;for(int i = 0; i < n; ++i) {int j = sa[rank[i] - 1];if(h > 0) --h;for(; j + h < n && i + h < n; ++h) {if(S[j + h] != S[i + h]) break;}lcp[rank[i] - 1] = h;}}char str[N];void gao() {build_sa(str, sa);build_lcp(str, sa, lcp);int ans = 0;for(int i = 0; i <= n; ++i) {ans += n - sa[i];if(i) ans -= lcp[i - 1];}printf("%d\n", ans);}int main() {int T;scanf("%d", &T);while(T--) {scanf("%s", str);gao();}return 0;}

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