Nightmare Ⅱ（双向BFS）

Problem Description

Last night, little erriyue had a horrible nightmare. He dreamed that he and his girl friend were trapped in a big maze separately. More terribly, there are two ghosts in the maze. They will kill the people. Now little erriyue wants to know if he could find his girl friend before the ghosts find them.
You may suppose that little erriyue and his girl friend can move in 4 directions. In each second, little erriyue can move 3 steps and his girl friend can move 1 step. The ghosts are evil, every second they will divide into several parts to occupy the grids within 2 steps to them until they occupy the whole maze. You can suppose that at every second the ghosts divide firstly then the little erriyue and his girl friend start to move, and if little erriyue or his girl friend arrive at a grid with a ghost, they will die.
Note: the new ghosts also can devide as the original ghost.

Input

The input starts with an integer T, means the number of test cases.
Each test case starts with a line contains two integers n and m, means the size of the maze. (1<n, m<800)
The next n lines describe the maze. Each line contains m characters. The characters may be:
‘.’ denotes an empty place, all can walk on.
‘X’ denotes a wall, only people can’t walk on.
‘M’ denotes little erriyue
‘G’ denotes the girl friend.
‘Z’ denotes the ghosts.
It is guaranteed that will contain exactly one letter M, one letter G and two letters Z.

Output

Output a single integer S in one line, denotes erriyue and his girlfriend will meet in the minimum time S if they can meet successfully, or output -1 denotes they failed to meet.

SampleInput

3
5 6
XXXXXX
XZ..ZX
XXXXXX
M.G...
......
5 6
XXXXXX
XZZ..X
XXXXXX
M.....
..G...

10 10
..........
..X.......
..M.X...X.
X.........
.X..X.X.X.
.........X
..XX....X.
X....G...X
...ZX.X...
...Z..X..X

SampleOutput

1
1
-1


题意就是给你一个迷宫，不对，就是说你现在被困在迷宫里，要去和你GF见面，但是迷宫中还有人不可以走的墙和会杀死人的幽灵，幽灵每个单位时间都会往上下左右延申新的幽灵。
幽灵有两个，M是你的位置，G是GF的位置。
幽灵每秒可以延申两格，你可以每秒走三步，GF每秒只能走一步，如果在不被杀死的情况和女朋友汇合就输出最小单位时间，否则输出-1.
（注意幽灵是可以往墙上延申的）
因为两个人都可以动，不用说，双向BFS是肯定的，不过有一点就是，如何实现每秒走三步以及判断是否被杀死呢？
答案是我也不知道。这道题目就留给各位自己思考。
=7=嘻嘻 不闹了。其实可以换个思路去思考，我们可以用三个bfs代替走三步，但是如何判断是否被杀死呢，其实这道题可以不需要判断是否被杀死，而是判断人物走到能否在幽灵延申到某个位置之前走到那个位置，用曼哈顿距离判断就行了。

提一下什么是曼哈顿距离
常用距离度量方法有十一种，而我们大部分时间只用到欧氏距离和曼哈顿距离。
设两个点的坐标(X1,Y1),(X2,Y2);
欧氏距离就是坐标的直线距离 = sqrt((X2 - X1)2+(Y2 - Y1)2)
而曼哈顿距离就是以欧式距离为斜边构造直角三角形的两直角边和 = |X2 - X1| + |Y2 - Y1|
为什么有这么多构造方式以及其区别，这篇博文就不详细介绍了。

值得一题的是，因为是双向搜索，所以需要开两个二维数组或是一个三维数组分别标记你或者GF是否走过。
还有就是代码BFS中的

int len = q[w].size();
while(len--)

因为我们不是一次就搜索完，我们的BFS仅仅只是做走一步的作用，所以只把当前已经存的点的下一点存入就行了。若是觉得难以理解可以替换成while(!q[w].empty)观察每一步的输出情况。

代码：

#include <iostream>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <sstream>
#include <iomanip>
#include <map>
#include <stack>
#include <deque>
#include <queue>
#include <vector>
#include <set>
#include <list>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <iterator>
#include <cmath>
#include <bitset>
#include <ctime>
#include <fstream>
#include <limits.h>
#include <numeric>

using namespace std;

#define F first
#define S second
#define mian main
#define ture true

#define MAXN 1000000+5
#define MOD 1000000007
#define PI (acos(-1.0))
#define EPS 1e-6
#define MMT(s) memset(s, 0, sizeof s)
typedef unsigned long long ull;
typedef long long ll;
typedef double db;
typedef long double ldb;
typedef stringstream sstm;
const int INF = 0x3f3f3f3f;

int fx[4][2]={1,0,-1,0,0,1,0,-1};
char mp[810][810];
int vis[2][810][810];
int gx,gy,mx,my,n,m,step;　　//记录坐标，这里的step指的是GF走的步数，应该理解成走了多少单位时间
pair<int,int>cur,z[2];　　//z用来记录幽灵位置
queue<pair<int,int> >q[2];    //分别记录你和GF的路径

bool check(pair<int,int> x){
    if(x.F < 0 || x.S < 0 || x.F >= n || x.S >= m || mp[x.F][x.S] == 'X')
        return false;
    if((abs(x.F-z[0].F)+abs(x.S-z[0].S)) <= 2*step || (abs(x.F-z[1].F)+abs(x.S-z[1].S)) <= 2*step)    //判断在幽灵延申到某个点之前是否能走到
        return false;
    return true;
}

int bfs(int w){
    pair<int,int>tp,next;
    int len = q[w].size();
    while(len--){　　//注意这里不是搜完，因为是多次搜索，只需要把当前步骤行进完就行了
        tp = q[w].front();
        q[w].pop();
        if(!check(tp)) continue;
        for(int i = 0; i < 4; i++){
            next.F = tp.F + fx[i][0];
            next.S = tp.S + fx[i][1];
            if(!check(next))
                continue;
            if(!vis[w][next.F][next.S]){
                if(vis[1-w][next.F][next.S])    //判断下一个点是否对方已经走过
                    return 1;
                vis[w][next.F][next.S] = 1;
                q[w].push(next);
            }
        }
    }
    return 0;
}

int solve(){
    while(!q[0].empty())
        q[0].pop();
    while(!q[1].empty())
        q[1].pop();

    cur.F = mx;
    cur.S = my;
    q[0].push(cur);
    cur.F = gx;
    cur.S = gy;
    q[1].push(cur);
    MMT(vis);
    vis[0][mx][my] = vis[1][gx][gy] = 1;
    step = 0;

    while((!q[0].empty()) || (!q[1].empty())){
        step++;
        if(bfs(0))    //通过三次bfs达到走三步
            return step;
        if(bfs(0))
            return step;
        if(bfs(0))
            return step;
        if(bfs(1))
            return step;
    }
    return -1;
}

int main(){
    ios_base::sync_with_stdio(false);
    cout.tie(0);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--){
        int cnt = 0;
        cin>>n>>m;
        for(int i = 0; i < n; i++)
            cin>>mp[i];
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++){
                if(mp[i][j] == 'G')
                    gx = i, gy = j;
                if(mp[i][j] == 'M')
                    mx = i, my = j;
                if(mp[i][j] == 'Z')
                    z[cnt].F = i, z[cnt++].S = j;
            }
        cout << solve() << endl;
    }
    return 0;
}