Tree array algorithm analysis

/*************************************** **************************************** * ***** Tree arrays include one-dimensional tree arrays and two-dimensional tree arrays; the main problem model is the known array a [n]. The subscript starts from 1 and changes the elements in array a. The sum between I and j in array a must be obtained; in the tree array, S [k] stores the sum of the number elements represented by the first 1 on the right of the binary representation of the forward k starting from k; that is, if lowbit is k, it indicates the number represented by the first 1 on the right. What is stored in S [k] is the sum of lowbit elements starting from a [k; s1 = A1S2 = A1 + A2S3 = A3S4 = A1 + A2 + A3 + A4S5 = A5S6 = A5 + A6S7 = A7S8 = A1 + A2 + A3 + A4 + A5 + A6 + A7 + a8 time complexity analysis: modify element: for modifying an element, If the I-th element is modified, you can directly change it in the S array. For example, the modified element is a [2]. so it affects only S [2], S [4], and S [8] elements in the c array. You only need to modify them one by one, the worst complexity of this process is less than O (logN); query element: for search, such as search sum [k], you only need to search for the number of 1 in the Binary Expression of k to obtain the final result. For example, you can search for sum [7]. The Binary Expression of 7 contains 3 1, that is, you need to search for 3 times, that is, sum [7] = S [7] + S [6] + S [4]; implementation process: Taking 7 as an example, the binary value is 0111; starting from the first 1 on the right; when the first 1 on the right exists, it is 7 and the binary value is 0111, that is, S [7]. Then, remove this 1 and get 6. the binary value is 0110, that is, S [6]. then, discard the used 1 and get 4. The binary value is 0100, that is, S [4]. that is, sum [7] = S [7] + S [6] + S [4]; **************************************** *** **************************************** */# Include <iostream> # include <cstdio> # include <cmath> # include <cstring> # include <algorithm> using namespace std; const int MAX = 50000; class Binary_Indexed_Tree {public: int *** s; int type; public: Binary_Indexed_Tree (int t );~ Binary_Indexed_Tree (); void clear (); int lowbit (int x) {return x & (-x) ;}; void modify (int x, int value ); void modify (int x, int y, int value); int sum (int x, int y) ;}; Binary_Indexed_Tree: Binary_Indexed_Tree (int t): type (t) {int I; s = new int * [MAX + 1]; // One-dimensional tree array if (type = 1) {for (I = 0; I <= MAX; I ++) s [I] = new int;} // two-dimensional tree array else if (type = 2) {for (I = 0; I <= MAX; I ++) s [I] = New int [MAX + 1] ;}} Binary_Indexed_Tree ::~ Binary_Indexed_Tree () {int I; for (I = 0; I <= MAX; I ++) delete [] s [I]; delete [] s;} void Binary_Indexed_Tree :: clear () {int I, j; for (I = 0; I <= MAX; I ++) {if (type = 1) s [I] [0] = 0; else {for (j = 0; j <= MAX; j ++) s [I] [j] = 0 ;}}} void Binary_Indexed_Tree: modify (int x, int value) {while (x <= MAX) {s [x] [0] + = value; x + = lowbit (x) ;}} void Binary_Indexed_Tree: modify (int x, int y, int value) {int t Emp = y; while (x <= MAX) {y = temp; while (y <= MAX) {s [x] [y] + = value; y = y + lowbit (y);} x = x + lowbit (x);} int Binary_Indexed_Tree: sum (int x) {int ans = 0; while (x> 0) {ans + = s [x] [0]; x-= lowbit (x);} return ans;} int Binary_Indexed_Tree: sum (int x, int y) {int ans = 0, temp = y; while (x> 0) {y = temp; while (y> 0) {ans + = s [x] [y]; y = y-lowbit (y);} x = x-lowbit (x);} return ans ;}/* **************************************** * ************ Appendix: HDU1541http: // acm.hdu.edu.cn/showproblem.php? Pid = 1541 question: how many stars are there in the lower left of a star ****************************** * *********************/int com [MAX], res [MAX], N; int lowbit (int x) {return x & (-x);} void modify (int pos, int val) {while (pos <= MAX) {com [pos] + = val; pos + = lowbit (pos) ;}} int quy (int x) {int sum = 0; while (x> 0) {sum + = com [x]; x-= lowbit (x);} return sum;} int main () {int x; while (scanf ("% d ", & N )! = EOF) {memset (com, 0, sizeof (com); memset (res, 0, sizeof (res); for (int I = 1; I <= N; ++ I) {scanf ("% d % * d", & x); x ++; res [quy (x)] ++; modify (x, 1) ;}for (int I = 0; I <N; ++ I) printf ("% d \ n", res [I]) ;}return 0 ;}