最少步数

最少步数

这有一个迷宫，有0~8行和0~8列：

 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

0表示道路，1表示墙。

现在输入一个道路的坐标作为起点，再如输入一个道路的坐标作为终点，问最少走几步才能从起点到达终点？

（注：一步是指从一坐标点走到其上下左右相邻坐标点，如：从（3，1）到（4,1）。）

23 1  5 73 1  6 7样例输出
1211
/*int binary_search(int a[],int l,int r,int key){    int m;    while(r - l > 1)    {    m = l + (r - 1)/2;    if(a[m] <= key)        l = m;        else        r = m;    }    if(a[l] == key)    return l;    else return -1;}*/#include <bits/stdc++.h>using namespace std;struct node{    int x;    int y;    int step;};queue<node> q;int flag[4][2] = {0,-1,0,1,-1,0,1,0};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 bfs(int a1,int a2,int b1,int b2){    int vis[9][9];    memset(vis,0,sizeof(vis));    int x,y;    node Nq;    Nq.x = a1;    Nq.y = a2;    Nq.step = 0;    q.push(Nq);    vis[a1][a2];    while(!q.empty())    {        Nq = q.front();        if(Nq.x == b1&&Nq.y == b2)        {            break;        }        q.pop();        for(int i=0;i<4;i++)        {            x = Nq.x + flag[i][0];            y = Nq.y + flag[i][1];            if(x >= 0 && x < 9 && y>=0&&y < 9 && vis[x][y] == 0&& Map[x][y] == 0)            {                node e ={x,y,Nq.step + 1};                q.push(e);                vis[x][y] = 1;            }        }    }    if(q.empty())        return -1;    while(!q.empty())        q.pop();    return Nq.step;}int main(){    int n;    cin>>n;    while(n--)    {        int a,b,c,d;        cin>>a>>b>>c>>d;        int l = bfs(a,b,c,d);        cout<<l<<endl;    }    return 0;}