poj2049--Finding Nemo

Finding Nemo

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 7317		Accepted: 1698

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

Source

Beijing 2004

英语翻译要人命啊，题意：感谢http://blog.csdn.net/wangjian8006/article/details/7997609

有一个迷宫，在迷宫中有墙与门
有m道墙，每一道墙表示为(x,y,d,t）
x,y表示墙的起始坐标
d为0即向右t个单位，都是墙
d为1即向上t个单位，都是墙
有n道门，每一道门表示为(x,y,d)
x,y表示门的起始坐标
d为0即向右一个单位表示门
d为1即向上一个单位表示门
再给出你起点的位置(f1,f2)，并保证这个点的位置不会再墙或者门中，为起点到(0,0)最少要穿过多少条门

注意：在迷宫中不只有门和墙，还有空地

#include 
#include 
int p[210][210][2] ; //0代表向右，1代表向上  //0代表空白，1代表门，2代表墙
int s[4][2] = { {-1,0},{1,0},{0,1},{0,-1} } ;
struct node{
    int x , y , num ;
}q[100000];
int r[210][210] ;
int main()
{
    int m , n , i , j , x , y , d , t ;
    while(scanf("%d %d", &m, &n)!=EOF)
    {
        if(m == -1 && n == -1)
            break;
        memset(p,0,sizeof(p));
        memset(r,0,sizeof(r));
        for(i = 0 ; i < m ; i++)
        {
            scanf("%d %d %d %d", &x, &y, &d, &t);
            if(d == 0)
            {
                for(j = x ; j < x + t ; j++)
                    p[j][y][d] = 2 ;
            }
            else
            {
                for(j = y ; j < y + t ; j++)
                    p[x][j][d] = 2 ;
            }
        }
        for(i = 0 ; i < n ; i++)
        {
            scanf("%d %d %d", &x, &y, &d);
            p[x][y][d] = 1 ;
        }
        double xx , yy ;
        scanf("%lf %lf", &xx, &yy);
        x = xx ;
        y = yy ;
        if(x == 0 || y == 0 || x >= 199 || y >= 199)
            printf("0\n");
        else
        {
            int low = 0 , top = 1 ;
            q[0].x = x ;
            q[0].y = y ;
            q[0].num = 0 ;
            r[x][y] = 1 ;
            int num = 9999999 ;
            while(low < top)
            {
                node k ;
                k = q[low++] ;
                if( (k.x == 0 || k.y == 0 || k.x >= 199 || k.y >= 199) && k.num < num )
                {
                    num = k.num ;
                    continue ;
                }
                if(k.num >= num)
                    continue ;
                int u , v ;
                for(i = 0 ; i < 4 ; i++)
                {
                    u = k.x + s[i][0] ;
                    v = k.y + s[i][1] ;
                    if(u < 0 || v < 0 || r[u][v] == 1)
                        continue ;
                    if(i == 0 && p[k.x][k.y][1] != 2 )
                    {
                        q[top].x = u ;
                        q[top].y = v ;
                        if(p[k.x][k.y][1] == 1)
                            q[top].num = k.num+1 ;
                        else
                            q[top].num = k.num ;
                        r[u][v] = 1 ;
                        top++ ;
                    }
                    else if(i == 1 && p[u][v][1] != 2 )
                    {
                        q[top].x = u ;
                        q[top].y = v ;
                        if(p[u][v][1] == 1)
                            q[top].num = k.num+1 ;
                        else
                            q[top].num = k.num ;
                        r[u][v] = 1 ;
                        top++ ;
                    }
                    else if(i == 2 && p[u][v][0] != 2)
                    {
                        q[top].x = u ;
                        q[top].y = v ;
                        if(p[u][v][0] == 1)
                            q[top].num = k.num+1 ;
                        else
                            q[top].num = k.num ;
                        r[u][v] = 1 ;
                        top++ ;
                    }
                    else if(i == 3 && p[k.x][k.y][0] != 2)
                    {
                        q[top].x = u ;
                        q[top].y = v ;
                        if(p[k.x][k.y][0] == 1)
                            q[top].num = k.num+1 ;
                        else
                            q[top].num = k.num ;
                        r[u][v] = 1 ;
                        top++ ;
                    }
                }
            }
            if(num == 9999999)
                printf("-1\n");
            else
                printf("%d\n", num);
        }
    }
    return 0;
}