HDU 5008 boring string problem (question B of Xi'an Network Competition)

HDU 5008 boring string Problem

Question Link

Idea: Construct a suffix array. the array of height can be used to pre-process the prefix and number of prefixes starting with each Lexicographic Order (in fact, to remove duplicate strings), and then search for the corresponding position at the second vertex each time, next, find the smallest SA [I ].

Code:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int MAXLEN = 100005;struct Suffix {char str[MAXLEN];int s[MAXLEN];int sa[MAXLEN], t[MAXLEN], t2[MAXLEN], c[MAXLEN], n;int rank[MAXLEN], height[MAXLEN];ll sum[MAXLEN];void build_sa(int m) {n++;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;}n--;}void getHeight() {int i, j, k = 0;for (i = 1; i <= n; i++) rank[sa[i]] = i;for (i = 0; i < n; i++) {if (k) k--;int j = sa[rank[i] - 1];while (s[i + k] == s[j + k]) k++;height[rank[i]] = k;}}void getsum() {for (int i = 1; i <= n; i++)sum[i] = sum[i - 1] + n - sa[i] - height[i];}void init() {n = strlen(str);for (int i = 0; i < n; i++)s[i] = str[i] - 'a' + 1;s[n] = 0;build_sa(27);getHeight();getsum();}void query(ll &ls, ll &rs, ll k) {int u = lower_bound(sum + 1, sum + n + 1, k) - sum;if (u == n + 1) {ls = 0; rs = 0;printf("%I64d %I64d\n", ls, rs);return;}int len = k - sum[u - 1] + height[u];int st = sa[u];for (int i = u; i > 1; i--) {if (height[i] < len) break;st = min(st, sa[i - 1]);}for (int i = u + 1; i <= n; i++) {if (height[i] < len) break;st = min(st, sa[i]);}ls = st + 1; rs = st + len;printf("%I64d %I64d\n", ls, rs);}void solve() {init();ll l = 0, r = 0, v;int q;scanf("%d", &q);while (q--) {scanf("%I64d", &v);ll k = (l^r^v) + 1;query(l, r, k);}}} gao;int main() {while (~scanf("%s", gao.str)) {gao.solve();}return 0;}

HDU 5008 boring string problem (question B of Xi'an Network Competition)