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Finding Nemo （poj 2049 超級蛋疼的bfs）

編輯：關於C++

Language: Finding Nemo Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 8372 Accepted: 1944

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

Sample Output

5
-1

Source

Beijing 2004

題意：二位坐標內告訴一些牆和門，兒子被困在裡面，父親在（0，0）處出發去救兒子，要求穿過的門數最少，輸出最少門數。

思路：我是把它轉化成了平常的二維地圖，先從（0，0）dfs走遍迷宮外的所有能到達的點並標記，然後從兒子所在地出發bfs，step記錄穿過了多少扇門，當走到迷宮外遇到dfs訪問過的點就表示出來了。今天做的幾個poj上的題怎麼都這麼蛋疼，實在無語了，實際測試數據有大於199的！！！具體去看discuss吧，我也是看了裡面的才過。。。

代碼：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
#pragma comment (linker,/STACK:102400000,102400000)
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf(%d, &n)
#define sff(a,b)    scanf(%d %d, &a, &b)
#define sfff(a,b,c) scanf(%d %d %d, &a, &b, &c)
#define pf          printf
#define DBG         pf(Hi
)
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1000;
const int MAXN = 2005;
const int MAXM = 200010;

struct Node
{
    int x,y,step;
    bool operator<(const Node &a)const
    {
        return step>a.step;
    }
};

int N,M;
int n,m,sx,sy;
int dir[4][2]={1,0,0,1,-1,0,0,-1};
int mp[maxn][maxn];
bool vis[maxn][maxn];
int door[maxn][maxn];

bool isok(int x,int y)
{
    if (x>=0&&x<=N&&y>=0&&y<=M) return true;
    return false;
}

void dfs(int x,int y)
{
    mp[x][y]=2;
    for (int i=0;i<4;i++)
    {
        int dx=x+dir[i][0];
        int dy=y+dir[i][1];
        if (isok(dx,dy)&&mp[dx][dy]==0&&!door[dx][dy])
            dfs(dx,dy);
    }
}

int bfs()
{
    Node st,now;
    memset(vis,false,sizeof(vis));
    st.x=sx,st.y=sy,st.step=0;
    vis[sx][sy]=true;
    priority_queueQ;
    Q.push(st);
    while (!Q.empty())
    {
        st=Q.top();Q.pop();
        if (mp[st.x][st.y]==2)
            return st.step;
        for (int i=0;i<4;i++)
        {
            now.x=st.x+dir[i][0];
            now.y=st.y+dir[i][1];
            if (isok(now.x,now.y)&&mp[now.x][now.y]!=1&&!vis[now.x][now.y])
            {
                now.step=st.step;
                if (door[now.x][now.y]) now.step++;
                vis[now.x][now.y]=true;
                Q.push(now);
            }
        }
    }
    return -1;
}

int main()
{
//#ifndef ONLINE_JUDGE
//    freopen(C:/Users/lyf/Desktop/IN.txt,r,stdin);
//#endif
    int i,j,x,y,d,l;
    double xx,yy;
    while (scanf(%d%d,&m,&n))
    {
        if (n==-1&&m==-1) break;
        memset(mp,0,sizeof(vis));
        memset(door,0,sizeof(door));
        N=M=0;
        for (i=0;i>xx>>yy;
        if (n==0 && m==0)
			printf (0
);
		else if (xx<0||yy<0||xx>199||yy>199)
			printf (0
);
		else
        {
            sx=((int)xx)*2+1;
            sy=((int)yy)*2+1;
            dfs(1,1);
            printf(%d
,bfs());
        }
    }
    return 0;
}