這道題關鍵的就是怎麼用佇列去儲存訪問的節點，用結構體，在用c++的佇列，其他的就是基本的bfs

別人的程式碼

#include<iostream>
#include<stdio.h>
#include<queue>
#include<cstring>
using namespace std;
int a[9][9]=//**迷宮地圖**//
{
	1,1,1,1,1,1,1,1,1,
    1,0,0,1,0,0,1,0,1,
    1,0,0,1,1,0,0,0,1,
    1,0,1,0,1,1,0,1,1,
    1,0,0,0,0,1,0,0,1,
    1,1,0,1,0,1,0,0,1,
    1,1,0,1,0,1,0,0,1,
    1,1,0,1,0,0,0,0,1,
    1,1,1,1,1,1,1,1,1
};
int b[4]={-1,1,0,0};//**往上下左右四個方向**//
int c[4]={0,0,-1,1};
int visit[9][9];
struct node
{
	int x,y;//**座標**//
    int step;//**步數**//
};
queue<node> Q;
int bfs(int x1,int y1,int x2,int y2)
{
	int i,s,t;
    node e={x1,y1,0};//**從(x1,y1)開始，步數首先定義為0**//
    Q.push(e);//**將可以走的方格都插入到佇列**//
    visit[x1][y1]=1;//**標記visit[x1][y1]曾經被訪問過**//
    while(!Q.empty())//如果佇列不為空**//
	{
		e=Q.front(); //**e設為佇列的第一個元素**//
        if(e.x==x2 && e.y==y2)//如果找到就跳出迴圈**//
		{
			break;
		}
        Q.pop();//**取出佇列的第一個元素**//
        for(i=0;i<4;i++)
		{
			s=e.x+b[i];//四個方向進行搜尋**//
            t=e.y+c[i];
            if(s>=0 && s<=8 && t>=0 && t<=8 && !visit[s][t] && a[s][t]==0)//**曾經訪問過這個格子或者這個格子不能繼續走**//
			{
				node e1={s,t,e.step+1};//**步數+1**//
                Q.push(e1);//**插入步數+1的元素**//
                visit[s][t]=1;//**標記visit[s][t]被訪問過**//
			}
		}
	}
	if(Q.empty())//**進入死衚衕，到達不了**//
	{
		return -1;
	}
    while(!Q.empty())//**如果佇列不為空**//
	{
		Q.pop();//**清空佇列**//
	}
    return e.step;//**返回最少步數**//
}
int main()
{
	int k,s,x1,x2,y1,y2;
    scanf("%d",&s);
    while(s--)
	{
		scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
        memset(visit,0,sizeof(visit));
        k=bfs(x1,y1,x2,y2);
        printf("%d\n",k);
	}
    return 0;
} 
 

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<vector> 
#include<string>
using namespace std;
typedef struct {
      int x;
      int y;
    }point;
vector<point> v;
int map[9][9]={
 1,1,1,1,1,1,1,1,1, 
 1,0,0,1,0,0,1,0,1,
 1,0,0,1,1,0,0,0,1,
 1,0,1,0,1,1,0,1,1,
 1,0,0,0,0,1,0,0,1,
 1,1,0,1,0,1,0,0,1,
 1,1,0,1,0,1,0,0,1,
 1,1,0,1,0,0,0,0,1,
 1,1,1,1,1,1,1,1,1
 };
int step[9][9];
int vis[9][9];
point p;
bool ok; 
int index;  
int start_x, start_y, end_x, end_y;
int bfs(int x, int y, int time)  // time 每次記錄當前的步數 
{
    if(x==end_x && y==end_y)
    {
      ok = true;
      return 0;
    }
    
    if(x-1>=0&&y>=0&&!vis[x-1][y]&&map[x-1][y]==0)  //依次搜尋 
    {
      step[x-1][y] = time+1;
      p.x = x-1;
      p.y = y;
      vis[x-1][y] = 1;
      v.push_back(p);
    }
    
    if(x>=0&&y+1<9&&!vis[x][y+1]&&map[x][y+1]==0)
    {
      step[x][y+1] = time+1;
      p.x = x;
      p.y = y+1;
      vis[x][y+1] = 1;
      v.push_back(p);
    }
    
    if(x+1<9&&!vis[x+1][y]&&map[x+1][y]==0)
    {
      step[x+1][y] = time+1;
      p.x = x+1;
      p.y = y;
      vis[x+1][y] = 1;
      v.push_back(p);
    }
     
    if(y-1>=0&&!vis[x][y-1]&&map[x][y-1]==0)
    {
      step[x][y-1] = time+1;
      p.x = x;
      p.y = y-1;
      vis[x][y-1] = 1;
      v.push_back(p);
    }
    
}
int main()
{

   int ncase;
   scanf("%d",&ncase);
    while(ncase--)
    {
    cin>>start_x>>start_y>>end_x>>end_y;
    //memset(step,0,sizeof(step));
    for(int i=0; i<9; i++)
    for(int j=0; j<9; j++)
    {
    step[i][j] = 0;
    vis[i][j] = 0;
    }
    p.x = start_x;
    p.y = start_y;
    v.clear(); 
    //step[p.x][p.y]=0;
    v.push_back(p);
    ok = false;
    index = 0;
    while(!ok)
    {
      bfs(v[index].x, v[index].y, step[v[index].x][v[index].y]);   
      index++;
    }                       
    cout<<step[end_x][end_y]<<endl;
    }
   system("pause");
   return 0;
}