Nightmare Ⅱ

Nightmare Ⅱ
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1788 Accepted Submission(s): 452

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

#include <iostream>#include <cstring>#include <cstdio>#include <queue>#include <cmath>#include <algorithm>#define N 1010using namespace std;char map[N][N];int t,n,m,step;int sx,sy,ex,ey;int z[3][3];int dir[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};struct node{    int x,y;};int check(int x,int y)//曼哈顿距离判断{    int x1=abs(x-z[0][0])+abs(y-z[0][1]);    int x2=abs(x-z[1][0])+abs(y-z[1][1]);    if((x1<=2*step)||(x2<=2*step))        return 0;    return 1;}queue<node>q[3];int bfs(int num, int k, char s, char e){    q[2] = q[num];    for(int t = 0; t<k; t++)    {        while(!q[2].empty())        {            node st = q[2].front();            q[2].pop();            q[num].pop();            node ed;            if(!check(st.x, st.y))    continue;            for(int i = 0; i<4; i++)            {                int nx = st.x + dir[i][0];                int ny = st.y + dir[i][1];                if(nx < 0 || ny < 0 || nx >= n || ny >= m || map[nx][ny] == 'X')    continue;                if(!check(nx, ny) || map[nx][ny] == s)    continue;                if(map[nx][ny] == e)    return 1;                map[nx][ny] = s;                ed.x = nx, ed.y = ny;                q[num].push(ed);            }        }        q[2] = q[num];    }    return 0;}void BFS(){    int i;    step=0;    for(i=0; i<3; i++)    {        while(!q[i].empty())            q[i].pop();    }    node a;    a.x=sx,a.y=sy;    q[0].push(a);    a.x=ex,a.y=ey;    q[1].push(a);    while(!q[0].empty()&&!q[1].empty())    {        step++;        if(bfs(0,3,'M','G')||bfs(1,1,'G','M'))        {            printf("%d\n",step);            return;        }    }    printf("-1\n");    return;}int main(){    int i,j,k;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        k=0;        for(i=0; i<n; i++)        {            scanf("%s",map[i]);            for(j=0; j<m; j++)            {                if(map[i][j]=='M')                    sx=i,sy=j;                else if(map[i][j]=='G')                    ex=i,ey=j;                else if(map[i][j]=='Z')                {                    z[k][0]=i,z[k][1]=j;                    k++;                }            }        }        BFS();    }    return 0;}