POJ:2049Finding Nemo(bfs+优先队列)

http://poj.org/problem?id=2049

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

题目大意：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。

 题目解析：

这个题忒坑，让它坑死了，首先这个地图可能完全由空格组成，所以要判断n==0&&m==0的状况，其次那个死孩子可能在地图外面，所以要判断孩子的位置。

解决方案：

如果坐标的位置不乘2的话，在给墙赋完值后，在给窗户赋值就把墙的值给覆盖了，所以坐标乘2进行处理。

刚开始对于孩子的位置坐标怎么处理没想明白，之后懂了，（int）x*2+1,(int)y*2+1是奇数，

而坐标乘2之后都是偶数，又因为孩子不在墙上与窗户上，所以（int）x*2+1,(int)y*2+1还是在原本孩子

呆的范围内。因为两墙之内最短的距离是2。

#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <queue>
using namespace std;
int map[1002][1002],v[1002][1002];
int n,m,maxx,maxy;
float tx,ty;
struct node
{
    int ans;
    int x,y;
    friend bool operator<(struct node a,struct node b)//按ans从小到大排序
    {
        return a.ans>b.ans;
    }
};
struct node t,f;
int jx[]= {1,-1,0,0};
int jy[]= {0,0,1,-1};
void bfs(int ww,int ee)
{
    priority_queue<node>q;
    memset(v,0,sizeof(v));
    t.x=ww;
    t.y=ee;
    t.ans=0;
    q.push(t);
    v[ww][ee]=1;
    while(!q.empty())
    {
        t=q.top();
        q.pop();
        if(t.x==0&&t.y==0)
        {
            printf("%d
",t.ans);
            return ;
        }
        for(int i=0; i<4; i++)
        {
            f.x=t.x+jx[i];
            f.y=t.y+jy[i];
            if(f.x>=0&&f.x<=maxx&&f.y>=0&&f.y<=maxy&&v[f.x][f.y]==0&&map[f.x][f.y]!=2)
            {
                if(map[f.x][f.y]==0)
                {
                    f.ans=t.ans;
                }
                else if(map[f.x][f.y]==1)
                {
                    f.ans=t.ans+1;
                }
                q.push(f);
                v[f.x][f.y]=1;
            }
        }
    }
    printf("-1
");
}
int main()
{
    int xx,yy,zz,ww;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        if(n==-1&&m==-1) break;
        maxx=-1;
        maxy=-1;
        memset(map,0,sizeof(map));
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d%d",&xx,&yy,&zz,&ww);
            if(zz==0)
            {
                for(int j=xx*2; j<=(xx+ww)*2; j++)
                {
                    map[j][yy*2]=2;//墙是2
                }
                if(maxx<(xx+ww)*2)
                    maxx=(xx+ww)*2;
                if(maxy<yy*2)
                    maxy=yy*2;
            }
            else if(zz==1)
            {
                for(int j=yy*2; j<=(yy+ww)*2; j++)
                {
                    map[xx*2][j]=2;
                }
                if(maxy<(yy+ww)*2)
                    maxy=(yy+ww)*2;
                if(maxx<xx*2)
                    maxx=xx*2;
            }
        }
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d%d",&xx,&yy,&zz);
            if(zz==0)
            {
                map[xx*2+1][yy*2]=1;//路是1
            }
            else if(zz==1)
            {
                map[xx*2][yy*2+1]=1;
            }
        }
        scanf("%f%f",&tx,&ty);
        if(!(tx>=0&&tx<=199&&ty>=0&&ty<=199))//他在迷宫之外
        {
            printf("0
");
        }
        else  if(!n&&!m)
            printf("0
");
        else
        {
            int ww=(int)tx*2+1;
            int ee=(int)ty*2+1;
            maxx+=10;
            maxy+=10;
            bfs(ww,ee);
        }

    }
    return 0;
}