【ybtoj】【BFS】【例题1】走迷宫

---

link

传送门
题目

---

解题思路

普通广搜，记录一下步数

---

Code

#include <iostream>
#include <cstdio>

using namespace std;

const int way[4][2] = {
    
    {
    
    -1, 0}, {
    
    1, 0}, {
    
    0, -1}, {
    
    0, 1}};
int n, a[1010][1010], s[1010][1010], v[1010][1010], q[1000100][2], sx, sy, ex, ey, h, t;

bool check(int x,int y) {
    
    
	return (x > 0 && x <= n && y > 0 && y <= n);
}

void BFS() {
    
    
	q[++t][0] = sx, q[t][1] = sy, v[sx][sy] = 1;
	while (h++ < t) {
    
    
		for (int i = 0; i < 4; i++) {
    
    
			int xx = q[h][0] + way[i][0];
			int yy = q[h][1] + way[i][1];
			if (check(xx, yy) && !v[xx][yy] && !a[xx][yy]) {
    
    
				q[++t][0] = xx, q[t][1] = yy, v[xx][yy] = 1;
				s[xx][yy] = s[q[h][0]][q[h][1]] + 1;
				if (xx == ex && yy == ey)
					return;
			}
		}
	}
}

int main() {
    
     
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
		{
    
    
			char c;
			cin >> c;
			a[i][j] = c - '0';
		}
	scanf("%d%d%d%d", &sx, &sy, &ex, &ey);
	BFS();
	printf("%d", s[ex][ey]);
}