【BFS】【YBTOJ】走迷宫

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思路：

bfs
判断0和1然后模拟
最后统计一下步数就好了

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#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;

const int dx[5] = {
    
    0, 0, 0, 1, -1};
const int dy[5] = {
    
    0, 1, -1, 0, 0};

int a[1005][1005], pd[1005][1005];
int dis[1005][1005];
int n, l, r;

bool check(int x, int y)
{
    
    
	if(x > n || y > n || x < 1 || y < 1)return 0;
	if(a[x][y] || pd[x][y])return 0;
	return 1;
}

void BFS(int a, int b)//搜索
{
    
    
	memset(pd, false, sizeof(pd));
	memset(dis, 0, sizeof(dis));
	queue<int>hx, hy;
	hx.push(a), hy.push(b);
	pd[a][b] = 1, dis[a][b] = 0;
	while(hx.size())
	{
    
    
		int xx = hx.front(); hx.pop();
		int yy = hy.front(); hy.pop();
		int x = xx + dx[1], y = yy + dy[1];
		for(int i = 1; i <= 4; i++)
			if(check(xx + dx[i], yy + dy[i]))
			{
    
    
				x = xx + dx[i], y = yy + dy[i];
				dis[x][y] = dis[xx][yy] + 1;
				pd[x][y] = 1;
				hx.push(x), hy.push(y);
				if(x == l && y == r)
				{
    
    
					printf("%d", dis[x][y]);
					return;
				}
			}
	}
	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';//转换
		}	
	int x, y;
	scanf("%d%d%d%d", &x, &y, &l, &r);
	BFS(x, y);
	return 0;
}