POJ-2155 Matrix (two-dimensional tree array + interval change + single-point evaluation or two-dimensional segment tree + interval update + single-point evaluation)

POJ-2155Matrix Time Limit: 3000MS Memory Limit: 65536KB 64bit IO Format: %i64d &%i64u Submit Status Description Given an n*n matrix A, whose elements is either 0 or 1. A[i, j] means the number in the I-th row and j-th column. Initially we have a[i, j] = 0 (1 <= i, J <= N). We can change the matrix in the following. Given a rectangle whose upper-left corner is (x1, y1) and Lower-right Corner are (x2, y2), we change all the elements in th e Rectangle by using the "not" operation (if it was a ' 0 ' then change it to ' 1 ' otherwise change it into ' 0 '). To maintain the information of the matrix, you is asked to write a program to receive and execute both kinds of instructio Ns. 1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= N, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whos E upper-left Corner is (x1, y1) and Lower-right Corner are (x2, y2). 2. Q x Y (1 <= x, y <= N) querys a[x, y]. Input The first line of the input was an integer X (x <=) representing the number of test cases. The following X blocks each represents a test case. The first line of each block contains numbers n and T (2 <= n <=, 1 <= T <= 50000) representing the S Ize of the Matrix and the number of the instructions. The following T lines each represents an instruction have the format "Q x y" or "C x1 y1 x2 y2", which has been describe D above. Output For each querying output one line, which have an integer representing A[x, y]. There is a blank line between every the continuous test cases. Sample Input 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1 Sample Output 1001 The title is to give you a matrix, the most beginning is 0, and then give you a top-left corner of the coordinates, the lower right corner of the coordinates, the area of 0, and 0, and then give you a point to let you find out whether he is the result of 1 or 0. You can use the interval changes in the tree array and the single point evaluation. Just here is a two-dimensional tree-like array, the principle is the same. If it changes, it can be so, if it starts [1....x2][1....y2] as an even number (this is considered a prefix and) first add [1....x2][1....y2] 1 to an odd number, then add [1...x1-1][y2] 1 to even, add [1....x2][y1] 1 to even, and they overlap [1....x1-1 ][1....X2-1] Part was changed three times. It's still odd, so there's a plus 1. Turn him into an even number, so that except [X1....x2][y1....y2] has changed. The rest of the place is still unchanged. /*author:2486memory:4304 kbtime:547 mslanguage:g++result:accepted*/#include <cstdio> #include <cstring> #include <algorithm>using namespace Std;const int maxn = $ + 5;int c[maxn][maxn];int N, X, M, x, y, x1, x2, y1, y 2;char op[10];int lowbits (int x) {return x & (-X);} void Add (int x,int y) {for (int i = x; I <= n;i + = lowbits (i)) {for (int j = Y;j <= n;j + = Lowbits (j)) { C[I][J] + +; }}}int query (int x,int y) {int ret = 0; for (int i = x;i > 0;i-= lowbits (i)) {for (int j = y;j > 0;j-= Lowbits (j)) {ret + = C[i][j]; }} return ret;} int main () {//freopen ("D://imput.txt", "R", stdin); scanf ("%d", &x); while (X-) {memset (c, 0, sizeof (c)); scanf ("%d%d", &n, &m); while (M-) {scanf ("%s", op); if (op[0] = = ' C ') {scanf ("%d%d%d%d", &x1, &y1, &x2, &y2); X2 + +; y2 + +; Add (x1, y1); Add (x1, y2); Add (x2, y1); Add (x2, y2); } else{scanf ("%d%d", &x, &y); printf ("%d\n", query (x, y) & 1);//sum represents the point of order} if (x) printf ("\ n"); } return 0;} Line segment Tree Code: /*author:2486memory:63688 kbtime:1579 mslanguage:g++result:accepted*/#include <cstdio> #include <cstring > #include <algorithm> #include <vector> #include <queue>using namespace std; #define Lson RT < < 1, l, Mid#define Rson RT << 1|1, Mid + 1, r#define root 1, 1, nconst int maxn = + 5;int N, T, case, X1, y1 , x2, y2, ret, x, Y;char op[5];int sum[maxn << 2][maxn << 2];void update_r (int cr, int y1, int y2, int rt, int l, int R) {if (Y1 <= l && r <= y2) {Sum[cr][rt] =!sum[cr][rt]; return; } int mid = (L + r) >> 1; if (y1 <= mid) Update_r (CR, Y1, Y2, Lson); if (Y2 > Mid) Update_r (CR, Y1, Y2, Rson);} void Update_c (int x1, int y1, int x2, int y2, int rt, int l, int r) {if (X1 <= l && r <= x2) {upd Ate_r (RT, y1, Y2, root); Return } int mid = (L + r) >> 1; if (x1 <= mid) Update_c (x1, y1, x2, y2, Lson); if (X2 > Mid) Update_c (x1, y1, x2, y2, Rson);} void Query_r (int cr, int y1, int y2, int rt, int l, int r) {if (Sum[cr][rt]) ret + +; if (Y1 <= l && r <= y2) {return; } int mid = (L + r) >> 1; if (y1 <= mid) Query_r (CR, Y1, Y2, Lson); if (Y2 > Mid) Query_r (CR, Y1, Y2, Rson);} void Query_c (int x1, int y1, int x2, int y2, int rt, int l, int r) {query_r (rt, y1, Y2, root);//Choose to overwrite the number of subtrees if (x1 &l t;= l && r <= x2) {return; } int mid = (L + r) >> 1; if (x1 <= mid) Query_c (x1, y1, x2, y2, Lson); if (X2 > Mid) Query_c (x1, y1, x2, y2, Rson);} int main () {//freopen ("D://imput.txt", "R", stdin); scanf ("%d", &case); while (case-) {memset (sum, 0, sizeof (sum)); scanf ("%d%d", &n, &t); while (T--) {scanf ("%s", op); if (op[0] = = ' C ') {scanf ("%d%d%d%d", &x1, &y1, &x2, &y2); Update_c (x1, y1, x2, y2, root); } else { scanf ("%d%d", &x, &y); ret = 0; Query_c (x, y, x, y, root); printf ("%d\n", ret & 1); }} if (case) printf ("\ n"); } return 0;}

POJ-2155 Matrix (two-dimensional tree array + interval change + single-point evaluation or two-dimensional segment tree + interval update + single-point evaluation)