HDU 4691 (ninth multi-school field 1006) Suffix Array

... What else can I say.

A drop of eyes slide over.

I always thought that the question is the LCP of the current string and all the strings previously entered... Then it's just a whole game.

A glance at the mark suddenly found that he only compared the input string with the previous string?

Suddenly, tens of millions of Grass-mud horses have been in my heart.

Then you can easily get...

Rmq is pre-processed first, and the complexity is nlogn. Then, each LCP query requires only the complexity of O (1.

# Include <set> # include <map> # include <stack> # include <cmath> # include <queue> # include <cstdio> # include <string> # include <vector> # include <iomanip> # include <cstring> # include <iostream> # include <algorithm> # define Max 2505 # define Fi first # define se second # define ll _ int64 # define pi ACOs (-1, 1.0) # define INF 0x3fffffff # define LL (x) (x <1) # define bug puts ("here") # define PII pair <int, int> # define RR (X) (X <1 | 1) # define MP (a, B) make_pair (a, B) # define MEM (a, B) memset (a, B, sizeof ()) # define rep (I, S, T) for (INT I = (s); I <= (t); ++ I) using namespace STD; inline void RD (Int & RET) {char C; int flag = 1; do {c = getchar (); If (C = '-') Flag =-1 ;} while (C <'0' | C> '9'); ret = C-'0'; while (C = getchar ()> = '0' & C <= '9') ret = RET * 10 + (c-'0'); RET * = flag;} inline void ot (int A) {if (a> = 10) ot (A/10); putchar (a % 10 + '0 ');} # define n 1000005/***** suffix array template *****/# define f (x) (X)/3 + (x) % 3 = 1? 0: Tb) // F (x) To find the starting position of suffix (x) of the original string in the new string # define g (x) <TB? (X) * 3 + 1 :( (x)-TB) * 3 + 2) // G (x) is the position of the new string suffix (x) in the original string, and f (x) are reciprocal operations int wa [N], WB [N], WV [N], WS [N]; int SA [N * 3]; int rank1 [N], height [N]; int R [N * 3]; int C0 (int * r, int A, int B) {return R [a] = R [B] & R [A + 1] = R [B + 1] & R [A + 2] = R [B + 2];} int C12 (int K, int * r, int A, int B) {If (k = 2) return R [a] <R [B] | (R [a] = R [B] & C12 (1, R, A + 1, B + 1); else return R [a] <R [B] | (R [a] = R [B] & W V [A + 1] <WV [B + 1]);} void sort (int * r, int * a, int * B, int N, int m) {int I; for (I = 0; I <n; I ++) WV [I] = R [A [I]; for (I = 0; I <m; I ++) WS [I] = 0; for (I = 0; I <n; I ++) WS [wv [I] ++; for (I = 1; I <m; I ++) WS [I] + = ws [I-1]; for (I = n-1; I> = 0; I --) B [-- ws [wv [I] = A [I]; return;} // Note: To facilitate recursive processing, the size of the r array and SA array is 3 * nvoid DC3 (int * r, int * Sa, int N, int m) {// The RN array stores the new recursive string, and the San array is the SA int I of the New String, J, * rn = R + N, * San = SA + N, TA = 0, TB = (n + 1)/3, TBC = 0, P; R [N] = R [n + 1] = 0; for (I = 0; I <n; I ++) {if (I % 3! = 0) wa [TBC ++] = I; // TBC indicates the number of suffixes where the modulo 3 is 1 or 2 at the starting position} Sort (R + 2, WA, WB, TBC, m); sort (R + 1, WB, WA, TBC, m); sort (R, WA, WB, TBC, m); For (P = 1, rn [F (WB [0])] = 0, I = 1; I <TBC; I ++) Rn [F (WB [I])] = C0 (r, WB [I-1], WB [I])? P-1: P ++; If (P <TBC) DC3 (RN, San, TBC, P); else {for (I = 0; I <TBC; I ++) San [Rn [I] = I ;}// sorts all suffixes where the modulo 3 is equal to 0 for (I = 0; I <TBC; I ++) {If (San [I] <TB) WB [ta ++] = San [I] * 3;} If (N % 3 = 1) // n % 3 = 1, special processing of suffix (n-1) WB [ta ++] = n-1; sort (R, WB, WA, Ta, M ); for (I = 0; I <TBC; I ++) WV [WB [I] = g (San [I])] = I; // merge the sorting results of all suffixes and save them in the SA array for (I = 0, j = 0, P = 0; I <TA & J <TBC; P ++) SA [p] = C12 (WB [J] % 3, R, wa [I], WB [J])? Wa [I ++]: WB [J ++]; for (; I <ta; P ++) SA [p] = wa [I ++]; (; j <TBC; P ++) SA [p] = WB [J ++]; return ;}// height [I] = suffix (SA [I-1]) the longest public prefix of suffix (SA [I]), that is, the longest public prefix of the two adjacent suffixes void calheight (int * r, int * Sa, int N) {int I, j, k = 0; for (I = 1; I <= N; I ++) rank1 [SA [I] = I; for (I = 0; I <n; height [rank1 [I ++] = k) for (K? K --: 0, j = sa [rank1 [I]-1]; R [I + k] = R [J + k]; k ++ );} int rmq [N]; int mm [N]; int best [20] [N]; void initrmq (int n) {int I, j, A, B; for (Mm [0] =-1, I = 1; I <= N; I ++) mm [I] = (I & (I-1 )) = 0 )? Mm [I-1] + 1: Mm [I-1]; for (I = 1; I <= N; I ++) best [0] [I] = I; for (I = 1; I <= mm [N]; I ++) for (j = 1; j <= n + 1-(1 <I); j ++) {A = Best [I-1] [J]; B = Best [I-1] [J + (1 <(I-1)]; If (rmq [a] <rmq [B]) best [I] [J] = A; else best [I] [J] = B;} return;} int askrmq (int A, int B) {int T; T = mm [B-A + 1]; B-= (1 <t)-1; A = Best [T] [a]; B = Best [T] [B]; return rmq [a] <rmq [B]? A: B;} int LCP (int A, int B) {int t; A = rank1 [a]; B = rank1 [B]; If (A> B) {T = A; A = B; B = T;} return (height [askrmq (a + 1, B)]);} /*************************************** * *****/# define n 1000005 char a [n]; int N; int CAL (INT now) {If (now = 0) return 1; int nn = 0; while (now) {NN ++; now/= 10 ;} return nn;} int main () {int TTT = 0; while (scanf ("% s", )! = EOF) {int L = strlen (a); For (INT I = 0; I <L; I ++) R [I] = A [I]; R [l] = 0; DC3 (R, SA, L + 1,200); calheight (R, SA, L); For (INT I = 1; I <= L; I ++) rmq [I] = height [I]; initrmq (l); rd (n); LL num1 = 0, num2 = 0; int X, Y; int prex =-1, prey =-1; while (n --) {RD (x); rd (y); num1 + = (Y-x) + 1; if (prex =-1) {num2 + = (Y-x) + 3; prex = x; prey = y; continue;} int now = 0; if (x = prex) {now = min (Prey-prex, Y-x);} else {now = LCP (prex, X ); now = min (Prey-prex, min (Y-X, now);} prex = x; prey = y; int FK = now; now = Y-X-now; if (now = 0) {num2 + = 2 + CAL (FK);} else num2 + = CAL (FK) + now + 2 ;} printf ("% i64d % i64d \ n", num1, num2); TTT ++;} return 0 ;}