Finding Nemo_BFS

Description

Nemo is a naughty boy. One day he went into the deep sea all by himself. Unfortunately, he became lost and couldn't find his way home. Therefore, he sent a signal to his father, Marlin, to ask for help. 
After checking the map, Marlin found that the sea is like a labyrinth with walls and doors. All the walls are parallel to the X-axis or to the Y-axis. The thickness of the walls are assumed to be zero. 
All the doors are opened on the walls and have a length of 1. Marlin cannot go through a wall unless there is a door on the wall. Because going through a door is dangerous (there may be some virulent medusas near the doors), Marlin wants to go through as few doors as he could to find Nemo. 
Figure-1 shows an example of the labyrinth and the path Marlin went through to find Nemo. 

We assume Marlin's initial position is at (0, 0). Given the position of Nemo and the configuration of walls and doors, please write a program to calculate the minimum number of doors Marlin has to go through in order to reach Nemo.

Input

Output

Sample Input

8 9

1 1 1 3

2 1 1 3

3 1 1 3

4 1 1 3

1 1 0 3

1 2 0 3

1 3 0 3

1 4 0 3

2 1 1

2 2 1

2 3 1

3 1 1

3 2 1

3 3 1

1 2 0

3 3 0

4 3 1

1.5 1.5

4 0

1 1 0 1

1 1 1 1

2 1 1 1

1 2 0 1

1.5 1.7

-1 -1

Sample Output

5

-1

#include <iostream>

#include<stdio.h>

#include<string.h>

#include<queue>

using namespace std;

const int inf=0x3f3f3f3f;

const int N=;

int xa[N][N],ya[N][N];
//xa記錄（i,j）的格子的上面的邊的狀態，ya記錄(i,j)的格子的右面的邊的狀態
//用inf表示牆，0表示空，1表示門；

int dis[N][N];

int di[][]={{,},{-,},{,},{,-}};

int n,m;

int wall=inf;

int maxx,maxy;

int get_val(int x,int y,int op)//檢驗一下有沒有穿越門

{

    if(op==) return ya[x][y];向右，檢測當前格子的右邊的值

    else if(op==) return ya[x-][y];//向左，檢測左邊的格子的右邊值

    else if(op==) return xa[x][y];//向上，檢測當前格子的上邊值

    else return xa[x][y-];//向下，檢測下面格子的上邊值

}

bool go(int x,int y)//檢測是否在範圍內

{

    if(x<||x>maxx||y<||y>maxy) return false;

    else return true;

}

int bfs(int sx,int sy)

{

    queue<int>qu;

    for(int i=;i<=maxx;i++)

    {

        for(int j=;j<=maxy;j++)

        {

            dis[i][j]=inf;

        }

    }

    dis[][]=;//從（1,1）這個格子出發

    qu.push();//橫縱座標都入隊

    qu.push();

    while(!qu.empty())

    {

        int nowx=qu.front();qu.pop();

        int nowy=qu.front();qu.pop();

        for(int i=;i<;i++)//向四個方向查詢

        {

            int xx=nowx+di[i][];

            int yy=nowy+di[i][];

            int tmp=get_val(nowx,nowy,i);

            if(go(xx,yy)&&dis[xx][yy]>dis[nowx][nowy]+tmp)//在範圍內，並經過的門少，則入隊

            {

                dis[xx][yy]=dis[nowx][nowy]+tmp;

                qu.push(xx);

                qu.push(yy);

            }

        }

    }

    return dis[sx][sy]==inf?-:dis[sx][sy];//inf表示此路不通，無路可走

}

int main()

{

    while(~scanf("%d%d",&n,&m))

    {

        if(m==-&&n==-) break;

        memset(xa,,sizeof(xa));

        memset(ya,,sizeof(ya));

        maxx=-,maxy=-;

        for(int i=; i<=n; i++)

        {

            int x,y,op,t;

            scanf("%d%d%d%d",&x,&y,&op,&t);

            if(op)

            {

                for(int j=; j<t; j++)

                    ya[x][y+j+]=wall;

                maxx=max(maxx,x+);

                maxy=max(maxy,y+t+);

            }

            else

            {

                for(int j=;j<t;j++)

                {

                    xa[x+j+][y]=wall;

                }

                maxx=max(maxx,x+t+);

                maxy=max(maxy,y+);

            }

        }

        for(int i=;i<=m;i++)

        {

            int x,y,op;

            scanf("%d%d%d",&x,&y,&op);

            if(op)

            {

                ya[x][y+]=;

            }

            else xa[x+][y]=;

        }

        double sx,sy;

        scanf("%lf%lf",&sx,&sy);

        if(sx<||sx>||sy<||sy>) printf("0\n");

        else printf("%d\n",bfs((int)sx+,(int)sy+));

    }

    return ;

}