Use the astar algorithm to solve the labyrinth shortest path problem

I wrote this maze for 7 rows, 7 columns, and 7 columns. The write is urgent. I just want to calculate the maze of N rows and M columns.

The Code is as follows:

# Include <iostream> # include <set> # include <string> # include <cmath> # include <cstdlib> # define n 100 using namespace STD; struct node {double H, g, F; int X, Y; bool operator <(node other) const {return F <Other. f ;}}; int map [N] [N]; // The Maze. 1 indicates that the path can be taken. 0 indicates that the path cannot go through int row or Col; // et <node> open; // you can expand the node Multiset <node>: iterator it; node close [N] [N]; // The node that has passed the int step [8] [2] = {,-, 0,-,-1,-,-1, -1}; // method I NT Prev [N * n]; // store the previous node void Init () {memset (close, 0, sizeof (close);} double getg (node, node B) // obtain the PATH value {return SQRT (1.0 * (. x-b.x) * (. x-b.x) +. y-b.y) * (. y-b.y);} bool is_match (node A, Node B) {if (. X = B. X &. y = B. y) {return true;} return false;} node begin, end; int main () {While (CIN> row> col) {open. clear (); Init (); For (INT I = 0; I <row; I ++) // input the maze {for (Int J = 0; j <Col; j ++) {CIN> map [I] [J] ;}} CIN> begin. x> B Egin. y> end. x> end. y; // start point and end point begin. G = getg (begin, end); // initialize the starting point of the Value Function begin. F = begin. g; begin. H = 0; open. insert (BEGIN); // bool issuccess = false; while (! Open. empty () // but there are still nodes that can go through {node son = * (open. begin (); // extracts the path consumption + the node open with the smallest valuation function. erase (open. begin (); // Delete the above node from the Open table for (INT I = 0; I <8; I ++) // search for 8 directions {int u = Son. X + step [I] [0]; int v = Son. Y + step [I] [1]; if (u> = 0 & V> = 0 & U <row & V <Col & map [u] [v]) // If the node is in the matrix and the value is 1, run {// to update the node information, including the path consumption and valuation function node TMP; TMP. X = u; TMP. y = V; TMP. G = getg (TMP, end); TMP. H = Son. H + 1; TMP. F = TMP. H + TMP. g; bool isinopen = false; // indicates whether the IF (u) = End. X & V = end. y) // if it is the end point, exit the loop {Prev [u * 7 + V] = Son. x * 7 + son. y; // record the previous node issuccess = true; // indicates that the path goto breakpoint is successfully found;} For (IT = open. begin (); it! = Open. end (); It ++) {// If the TMP point is in the Open table, compare it with the F value in the Open table, if the former F is small, it inserts TMP into the Open table if (is_match (* It, TMP) {isinopen = true; // indicates if (* It) in the Open table ). f> TMP. f) {node new = TMP; open. erase (it); open. insert (TMP); Prev [TMP. x * 7 + TMP. y] = Son. x * 7 + son. y; // record the previous node break; }}} bool isinclose = false; // mark whether it is in the close table if (close [TMP. x] [TMP. y]. f) // if it is in close, it indicates that it has passed through and then compares the size of F. If the former F is small, insert the vertex into the Open table and delete the vertex from the close table. {Isinclose = true; // If (TMP. F <close [TMP. x] [TMP. y]. f) {open. insert (TMP); close [TMP. x] [TMP. y]. f = 0; close [TMP. x] [TMP. y]. G = 0; close [TMP. x] [TMP. y]. H = 0; Prev [TMP. x * 7 + TMP. y] = Son. x * 7 + son. y; // record the previous node} If (! Isinopen &&! Isinclose) // If the node is neither in the close table nor in the Open table, it indicates that the node has not passed and is inserted into the Open table. {Open. insert (TMP); Prev [TMP. x * 7 + TMP. y] = Son. x * 7 + son. y; // record the previous node }}// put the point retrieved from the Open table into the close table, which indicates that the node has passed through the close [son. x] [son. y]. X = Son. x; close [son. x] [son. y]. y = Son. y; close [son. x] [son. y]. G = Son. g; close [son. x] [son. y]. F = Son. f; close [son. x] [son. y]. H = Son. h;} breakpoint:; If (issuccess) // if the path is found, the output path is not output in reverse order, so the point is inverted. {Cout <"The path is:" <Endl; int x = end. x * 7 + end. y; int y = begin. x * 7 + begin. y; while (X! = Y) {cout <"(" <X/7 <"," <X % 7 <") --> "; X = Prev [X];} cout <"(" <begin. x <"," <begin. Y <")" <Endl; cout <"reach success! "<Endl;} else // cannot reach the end {cout <" reach failed! "<Endl ;}}}