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

The input consists of several test cases. Each test case is started by two non-negative integers M and N. M represents the number of walls in the labyrinth and N represents the number of doors. 
Then follow M lines, each containing four integers that describe a wall in the following format: 
x y d t 
(x, y) indicates the lower-left point of the wall, d is the direction of the wall -- 0 means it‘s parallel to the X-axis and 1 means that it‘s parallel to the Y-axis, and t gives the length of the wall. 
The coordinates of two ends of any wall will be in the range of [1,199]. 
Then there are N lines that give the description of the doors: 
x y d 
x, y, d have the same meaning as the walls. As the doors have fixed length of 1, t is omitted. 
The last line of each case contains two positive float numbers: 
f1 f2 
(f1, f2) gives the position of Nemo. And it will not lie within any wall or door. 
A test case of M = -1 and N = -1 indicates the end of input, and should not be processed.

Output

For each test case, in a separate line, please output the minimum number of doors Marlin has to go through in order to rescue his son. If he can‘t reach Nemo, output -1.

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

【题意】给出n，m分别表示墙数和门数

再给出n组x,y,op,t；op=1时表示起点为x,y的墙向上t个单位，op=0则表示起点为x,y的墙向右t个单位

再给出m组x,y,op；op=1时表示起点为x,y的门向上1个单位，op=0则表示起点为x,y的门向右1个单位

给出sx,sy；求从（1,1）到（sx，sy）最少经过的门

【思路】用xa记录（i,j）的格子的上面的边的状态，ya记录(i,j)的格子的右面的边的状态。进行bfs；

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int inf=0x3f3f3f3f;
const int N=210;
int xa[N][N],ya[N][N];//xa记录（i,j）的格子的上面的边的状态，ya记录(i,j)的格子的右面的边的状态//用inf表示墙，0表示空，1表示门；
int dis[N][N];
int di[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int n,m;
int wall=inf;
int maxx,maxy;
int get_val(int x,int y,int op)//检验一下有没有穿越门
{
    if(op==0) return ya[x][y];向右，检测当前格子的右边的值
    else if(op==1) return ya[x-1][y];//向左，检测左边的格子的右边值
    else if(op==2) return xa[x][y];//向上，检测当前格子的上边值
    else return xa[x][y-1];//向下，检测下面格子的上边值
}
bool go(int x,int y)//检测是否在范围内
{
    if(x<1||x>maxx||y<1||y>maxy) return false;
    else return true;
}
int bfs(int sx,int sy)
{
    queue<int>qu;
    for(int i=0;i<=maxx;i++)
    {
        for(int j=0;j<=maxy;j++)
        {
            dis[i][j]=inf;
        }
    }
    dis[1][1]=0;//从（1,1）这个格子出发
    qu.push(1);//横纵坐标都入队
    qu.push(1);
    while(!qu.empty())
    {
        int nowx=qu.front();qu.pop();
        int nowy=qu.front();qu.pop();
        for(int i=0;i<4;i++)//向四个方向查找
        {
            int xx=nowx+di[i][0];
            int yy=nowy+di[i][1];
            int tmp=get_val(nowx,nowy,i);
            if(go(xx,yy)&&dis[xx][yy]>dis[nowx][nowy]+tmp)//在范围内，并经过的门少，则入队
            {
                dis[xx][yy]=dis[nowx][nowy]+tmp;
                qu.push(xx);
                qu.push(yy);
            }
        }
    }
    return dis[sx][sy]==inf?-1:dis[sx][sy];//inf表示此路不通，无路可走
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(m==-1&&n==-1) break;
        memset(xa,0,sizeof(xa));
        memset(ya,0,sizeof(ya));
        maxx=-1,maxy=-1;
        for(int i=1; i<=n; i++)
        {
            int x,y,op,t;
            scanf("%d%d%d%d",&x,&y,&op,&t);
            if(op)
            {
                for(int j=0; j<t; j++)
                    ya[x][y+j+1]=wall;
                maxx=max(maxx,x+1);
                maxy=max(maxy,y+t+1);

            }
            else
            {
                for(int j=0;j<t;j++)
                {
                    xa[x+j+1][y]=wall;
                }
                maxx=max(maxx,x+t+1);
                maxy=max(maxy,y+1);
            }
        }
        for(int i=1;i<=m;i++)
        {
            int x,y,op;
            scanf("%d%d%d",&x,&y,&op);
            if(op)
            {
                ya[x][y+1]=1;
            }
            else xa[x+1][y]=1;
        }
        double sx,sy;
        scanf("%lf%lf",&sx,&sy);
        if(sx<1||sx>199||sy<1||sy>199) printf("0\n");
        else printf("%d\n",bfs((int)sx+1,(int)sy+1));
    }
    return 0;
}