HDU 3085 Nightmare Ⅱ(初学双向BFS)

题面

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.

Sample Input

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

Sample Output

1
1
-1

题目链接

HDU_3085

参考链接

HDU3085 Author:Salvete

题意

我和我的女朋友被困迷宫，我和我女朋友不可以穿墙，鬼可以穿墙。

我每秒走3步，女朋友每秒走1步，鬼每秒向上下左右扩展两个。即：

		Z

 

		Z
		Z
Z	Z	Z	Z	Z
		Z
		Z

 问我和女朋友最少多长时间（在不被鬼抓到的情况下）可以相遇？

分析

分别以我和女朋友为起点，进行bfs，为了模拟走一步，则对bfs搜索树，进行一层的遍历。

一秒走三步，则在一秒内，遍历3层。

程序说明

bool is_kill(Node now)
判断now节点是否合法，是否会被鬼杀掉
bool bfs(int person)/**I:person==0  Girlfriend:person==1**/
person走一步，对此时q进行一层遍历（层次遍历）
int work()
计算结果
void read()
初始化+读入

程序

#include<stdio.h>
#include<queue>
#include<math.h>
#include<cmath>
#include<string.h>
using namespace std;
#define maxn 800
int time;
struct Node
{
    int x,y;
    Node(int X,int Y):x(X),y(Y){}
    Node(){}
    void change(int X,int Y){x=X,y=Y;}
}I,girl,ghost[2];

queue<Node>q[2];

int zx[]={0,0,1,-1};
int zy[]={1,-1,0,0};
bool visit[2][maxn][maxn];
char MAP[maxn][maxn];
int n,m;

bool is_kill(Node now)
{
    if(now.x<0||now.x>=n||now.y<0||now.y>=m||MAP[now.x][now.y]=='X')
        return true;
    for(int i=0;i<=1;i++)
    {
        if(abs(now.x-ghost[i].x)+abs(now.y-ghost[i].y)<=time*2)
            return true;
    }
    return false;
}

bool bfs(int person)/**I:person==0  Girlfriend:person==1**/
{
    Node now,next;
    int Size=q[person].size();
    for(int i=0;i<Size;i++)
    {
        now=q[person].front();
        q[person].pop();
        if(is_kill(now))
            continue;
        for(int j=0;j<4;j++)
        {
            next=now;
            next.x+=zx[j];
            next.y+=zy[j];
            if((!is_kill(next))&&!visit[person][next.x][next.y])
            {
                if(visit[1-person][next.x][next.y])
                    return true;
                visit[person][next.x][next.y]=true;
                q[person].push(next);
            }
        }
    }
    return false;
}

int work()
{
    for(int i=0;i<2;i++)
        while(!q[i].empty())
            q[i].pop();
    memset(visit,0,sizeof(visit));
    q[0].push(I);
    q[1].push(girl);
    visit[0][I.x][I.y]=visit[1][girl.x][girl.y]=true;
    time=0;
    while(!q[0].empty()||!q[1].empty())
    {
        time++;
        if(bfs(0))
            return time;
//        printf("time=%d\n",time);
        if(bfs(0))
            return time;
//        printf("time=%d\n",time);
        if(bfs(0))
            return time;
//        printf("time=%d\n",time);
        if(bfs(1))
            return time;
    }
    return -1;
}

void read()
{
    memset(MAP,'\0',sizeof(MAP));
    scanf("%d%d",&n,&m);
    ghost[0].change(-1,-1);
    ghost[1].change(-1,-1);
    for(int i=0;i<n;i++)
    {
        scanf("%s",MAP[i]);
        for(int j=0;j<m;j++)
        {
            if(MAP[i][j]=='M')
                I.change(i,j);
            else if(MAP[i][j]=='G')
                girl.change(i,j);
            else if(MAP[i][j]=='Z')
            {
                if(ghost[0].x==-1)
                    ghost[0].change(i,j);
                else
                    ghost[1].change(i,j);
            }
        }
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        read();
        printf("%d\n",work());
    }
    return 0;
}