POJ 3984 Maze problem

Do this problem actually I want to see the way of printing path, water, look at the code of others on the internet, the print path is basically to save the parent node, and then

Print out its coordinates, I wrote a recursive, from behind to find the starting point, the process is a little different, not saved before, but continue up and down around the

Look, estimate is because this time a bit more, although in this problem can not see the difference ... Inadvertently saw the blog, I do this problem nearly a year later than her ...

Code:

#include <stdio.h> #include <string.h> #include <stdlib.h> #include <queue> int dist[10][10];
int visit[10][10];
int gra[10][10];
const int Dx[4] = { -1,1,0,0};
const int Dy[4] = {0,0,-1,1};
using namespace Std;
struct node {int x, y;};
	void BFs (int x,int y) {int i;
	queue<node>que;
	memset (visit,0,sizeof (visit));
	memset (dist,0,sizeof (Dist));
	Node N1;
	n1.x = x;
	N1.y = y;
	Dist[x][y] = 0;
	Que.push (N1);
	Visit[x][y] = 1;
		while (!que.empty ()) {node N1 = Que.front ();
		Que.pop ();
			if (n1.x = = 5) && (n1.y = = 5)) {//printf ("%d\n", Dist[n1.x][n1.y]);
		return;
			} for (i=0; i<4; i++) {int xx = n1.x + dx[i];
			int yy = N1.y + dy[i]; if (!visit[xx][yy] && (gra[xx][yy] = = 0) && (xx >= 1) && (xx <= 5) && (yy >= 1) &am
				p;& (yy <= 5)) {Visit[xx][yy] = 1;
				DIST[XX][YY] = dist[n1.x][n1.y] + 1;
				Node N2;
				n2.x = Xx,n2.y = yy;
			Que.push (N2); }}}} void Print_path(Node N1,node n2)
	{int i;
	if (n2.x = = n1.x) && (n1.y = = n2.y)) return;
		for (i=0; i<4; i++) {int xx = n2.x + dx[i];
		int yy = n2.y + dy[i];
			if ((dist[n2.x][n2.y] = = Dist[xx][yy] + 1) && Visit[xx][yy]) {node N3;
			n3.x =xx,n3.y =yy;
			Print_path (N1,N3);
		printf ("(%d,%d) \ n", n3.x-1,n3.y-1);
}} return;
	} int main () {int i,j;
	for (I=1; i<=5; i++) for (j=1; j<=5; j + +) scanf ("%d", &gra[i][j]);	
	BFS (a);
	Node n1,n2;
	n1.x = 1, n1.y = 1;
	n2.x = 5, n2.y = 5;
	Print_path (N1,N2);
	printf ("(4, 4)");
return 0;
 }