hdu4456 Crowd (two-dimensional tree-like array)

Test instructions: Gives a n*n matrix, and then M operation,1 represents the coordinates (x, y) value plus z,2 represents the weights and the points with coordinates (x, y) of Manhattan that are not more than Z.

Problem-solving ideas: The matrix side over 45 degrees. When a query is found, the effective point is actually a matrix.

Then it becomes a single point of change. Ask for the matrix and the problem.

We consider the practice of a bare two-dimensional tree-like array. It will be found that the matrix is too large, but notice that the initial time, all the values in the matrix are 0, then this two-dimensional tree array. The effective point is the point at which the change is made, and the interval in which these points are controlled. The total number of States here is only M*logn*logn. So we take the operation out and map the changed state to the sequence.

Code:

#include <stdio.h> #include <string.h> #include <algorithm> #include <vector> #define LOWBIT (x) (x& (-X)) using namespace std; struct Operation {int op, x, Y, Z;} p[88888]; int base = 100000; const int N = 22  22222;vector<int> MP; int val[n], tot, N, m; const int MAXN = 22222; inline int get (int x, int y) {x = x    * base + y;    y = Lower_bound (Mp.begin (), Mp.end (), X)-Mp.begin ();    if (y = = Tot | | mp[y]! = x) return-1; return y;}        void update (int x, int y, int z) {while (X < n*2+100) {int pos = y;            while (Pos < n*2+100) {Int. now = Get (X,pos);            Val[now] + = Z;        pos + = Lowbit (POS);    } x + = Lowbit (x);    }}int sum (int x, int y) {int ret = 0;        while (x) {int pos = y;            while (POS) {Int. now = Get (x, POS);            if (now! =-1) ret + = Val[now];        pos-= Lowbit (POS); } X-= Lowbit (x); } return ret;}  int query (int x1, int y1, int x2, int y2) {int ret1 = SUM (x1-1,y1-1), Ret2 = SUM (x2,y2), Ret3 = SUM (x1-1,y2),    Ret4 = SUM (x2,y1-1);//printf ("Ret2 =%d\n", Ret2); return RET1+RET2-RET3-RET4;}    int Get_num () {int n = 0, flag = 1;    char c; while ((c = GetChar ()) && c! = '-' && (c< ' 0 ' | |    C> ' 9 '));    if (c = = '-') flag =-1;    else n = C-' 0 ';    while ((c = GetChar ()) && C >= ' 0 ' && C <= ' 9 ') n = n * + C-' 0 '; return n * FLAG;            void Add (int t, int x, int y) {while (X < n*2+100) {int pos = y;                while (Pos < n*2+100) {//if (tot = = N-1) while (1);                Mp.push_back (x * base + POS);            pos + = Lowbit (POS);        } x + = Lowbit (x);    }}void read (int i) {p[i].op = Get_num ();    p[i].x = Get_num ();    P[I].Y = Get_num ();    P[i].z = Get_num (); Intx = p[i].x + p[i].y-1, y = n-p[i].x + p[i].y; if (P[i].op = = 1) Add (1, x, y);}        int main () {while (scanf ("%d", &n) && N) {scanf ("%d", &m);        Mp.clear ();        Memset (val, 0, sizeof (Val));        for (int i = 1; I <= m; i + +) read (i);        Sort (Mp.begin (), Mp.end ());        Vector<int>::iterator it = unique (Mp.begin (), Mp.end ());   Mp.erase (It,mp.end ());        Tot = Unique (MP + 1, MP + tot + 1)-mp-1; tot = Mp.size ();            for (int i = 1; I <= m; i + +) {int op, x, Y, Z;            op = p[i].op;            x = p[i].x;            y = p[i].y;    z = p[i].z;            printf ("Op =%d, x =%d, y =%d, z =%d\n", op, x, Y, p[i].z);            int xx = x + y-1, yy = N-x + y;            if (op = = 1) {update (xx, yy, z);      } else {int x1 = Std::max (1, xx-z);          int y1 = Std::max (1, yy-z);                int x2 = std::min (2*n-1, XX + z);                int y2 = std::min (2*n-1, yy + z);            printf ("%d\n", Query (x1, y1, x2, y2)); }}} return 0;}

hdu4456 Crowd (two-dimensional tree-like array)