海底总动员。。。。

这个题开始不会建图，彻底颠覆以前我对广搜题的想法。想了好久， 忽然想到省赛时HYPO让我做三维BFS来着，一直没做，看到POJ计划这个题，就是三维BFS解题，就做了一下， 对于这个题。。。。实在不知道说什么好，又坑、又SB，POJ的后台数据和题目描述的完全不一样，看了DIscuss之后开始 改动代码，最后改的又臭又长，搜了无数题解找数据，卡了整整两天。

挥挥洒洒 160行。。。。同时也是我第一次使用  三维建图+BFS，纪念一下！

2049 算是我攻克POJ计划的第一个卡两天的难关，POJ完成进度  10/200，为了庆贺一下，我决定回去睡个觉；

input

0 0
-100.0 -100.0

-1 -1

打印

0

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

Source

ME             time          WA     RE     AC

1536KB      32ms         15   +   9   =  1 

我的第三维只是储存 墙和门

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#define right 0
#define up    1
#define wall  3
#define door  4
#define null  0
const int N = 210;
const int Size = 999999;
const int INF = 1<<20;
using namespace std;
struct node{
    int x,y,ans;
}q[Size];
int mapp[N][N][2];
int vis[N][N];
int n,m,result;
int mv[4][2] = {{1,0},{0,-1},{0,1},{-1,0}};

void init()
{
    result = 0;
    memset(mapp,0,sizeof(mapp));//初始化空地
    memset(vis,0,sizeof(vis));
}
void BFS(int x,int y)
{
    node f,t;
    int s = 0,e = 0;
    f.x = x;
    f.y = y;
    f.ans = 0;
    q[e++] = f;
    vis[x][y] = 1;
    result = INF;
    while(s < e)
    {
        t = q[s++];
        if(t.x ==0 || t.y ==0 || t.x > 198 || t.y > 198 ) //边界处理，但是不能跳出，只能跳过，因为存在多个门的情况
        {
            if(result > t.ans)  //走出去之后，保存最小通过门数
                result = t.ans;
            continue; //注意，不要break，
        }

        for(int i = 0;i<4;i++)
        {
            f.x = t.x + mv[i][0];
            f.y = t.y + mv[i][1];
           // printf("asd\n");
            if(i==0)
            {
                if(!vis[f.x][f.y] && mapp[t.x][t.y][up]!=wall)//右
                {
                    if(mapp[t.x][t.y][up]==door)
                        f.ans = t.ans +1;
                    else
                        f.ans = t.ans;
                    vis[f.x][f.y] = 1;
                    q[e++] = f;
                }
            }
            else if(i==1)
            {
                if(!vis[f.x][f.y] && mapp[f.x][f.y][right]!=wall) //下
                {
                    if(mapp[f.x][f.y][right]==door)
                        f.ans = t.ans +1;
                    else
                        f.ans = t.ans;
                    vis[f.x][f.y] = 1;
                    q[e++] = f;
                }
            }
            else if(i==2)
            {
                if(!vis[f.x][f.y] && mapp[t.x][t.y][right]!=wall)//上
                {
                    if(mapp[t.x][t.y][right]==door)
                        f.ans = t.ans +1;
                    else
                        f.ans = t.ans;
                    vis[f.x][f.y] = 1;
                    q[e++] = f;
                }
            }
            else if(i==3)
            {
                if(!vis[f.x][f.y] && mapp[f.x][f.y][up]!=wall)//左
                {
                    if(mapp[f.x][f.y][up]==door)
                        f.ans = t.ans +1;
                    else
                        f.ans = t.ans;
                    vis[f.x][f.y] = 1;
                    q[e++] = f;
                }
            }
        }
    }
}

int main()
{
    int x,y,d,t;
    double Nemo_x,Nemo_y;
   while(~scanf("%d%d",&m,&n))
   {
       if(n==-1 && m==-1)
        break;
        init();
        for(int num_wall = 0;num_wall < m;num_wall++)
        {
            scanf("%d%d%d%d",&x,&y,&d,&t);
            if(d)
            {
                for(int num = 0;num < t;num++)
                     mapp[x-1][y+num][up] = wall;
            }
            else
            {
                for(int num = 0;num < t;num++)
                    mapp[x+num][y-1][right] = wall;
            }
        }
        for(int num_door = 0;num_door < n;num_door++)
        {
            scanf("%d%d%d",&x,&y,&d);
            if(d)
                mapp[x-1][y][up] = door;
            else
                mapp[x][y-1][right] = door;
        }
        int xx,yy;
        cin>>Nemo_x>>Nemo_y;
        xx = (int)(Nemo_x + 0.0001);//这里要注意 double 取整时 会有损失0.99999 会变成0
        yy = (int)(Nemo_y + 0.0001);
        if(n==0 && m==0)
       {
           printf("0\n");
           continue;
       }
        if(xx <=0 || yy <= 0 || xx >= 199 || yy >= 199)  //后台可能出现 0,0 200+,200+，虽然不符合题目描述
        {
            printf("0\n");

        }
        else
        {
            BFS(xx,yy);
        (result==INF)?printf("-1\n"):printf("%d\n",result);
}
   }
    return 0;
}