hdu 1072

Nightmare

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10962    Accepted Submission(s): 5331

Problem Description

Ignatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb explodes. The initial exploding time of the bomb is set to 6 minutes. To prevent the bomb from exploding by shake, Ignatius had to move slowly, that is to move from one area to the nearest area(that is, if Ignatius stands on (x,y) now, he could only on (x+1,y), (x-1,y), (x,y+1), or (x,y-1) in the next minute) takes him 1 minute. Some area in the labyrinth contains a Bomb-Reset-Equipment. They could reset the exploding time to 6 minutes.

Given the layout of the labyrinth and Ignatius' start position, please tell Ignatius whether he could get out of the labyrinth, if he could, output the minimum time that he has to use to find the exit of the labyrinth, else output -1.

Here are some rules:
1. We can assume the labyrinth is a 2 array.
2. Each minute, Ignatius could only get to one of the nearest area, and he should not walk out of the border, of course he could not walk on a wall, too.
3. If Ignatius get to the exit when the exploding time turns to 0, he can't get out of the labyrinth.
4. If Ignatius get to the area which contains Bomb-Rest-Equipment when the exploding time turns to 0, he can't use the equipment to reset the bomb.
5. A Bomb-Reset-Equipment can be used as many times as you wish, if it is needed, Ignatius can get to any areas in the labyrinth as many times as you wish.
6. The time to reset the exploding time can be ignore, in other words, if Ignatius get to an area which contain Bomb-Rest-Equipment, and the exploding time is larger than 0, the exploding time would be reset to 6.

 

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case starts with two integers N and M(1<=N,Mm=8) which indicate the size of the labyrinth. Then N lines follow, each line contains M integers. The array indicates the layout of the labyrinth.
There are five integers which indicate the different type of area in the labyrinth:
0: The area is a wall, Ignatius should not walk on it.
1: The area contains nothing, Ignatius can walk on it.
2: Ignatius' start position, Ignatius starts his escape from this position.
3: The exit of the labyrinth, Ignatius' target position.
4: The area contains a Bomb-Reset-Equipment, Ignatius can delay the exploding time by walking to these areas.

 

Output

For each test case, if Ignatius can get out of the labyrinth, you should output the minimum time he needs, else you should just output -1.

 

Sample Input

3
3 3
2 1 1
1 1 0
1 1 3
4 8
2 1 1 0 1 1 1 0
1 0 4 1 1 0 4 1
1 0 0 0 0 0 0 1
1 1 1 4 1 1 1 3
5 8
1 2 1 1 1 1 1 4 
1 0 0 0 1 0 0 1 
1 4 1 0 1 1 0 1 
1 0 0 0 0 3 0 1 
1 1 4 1 1 1 1 1 

 

Sample Output

4
-1
13

 

Author

Ignatius.L

 

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Ø 题目描述

    Ignatius昨晚做了一个恶梦，梦见在一个带有定时炸弹的迷宫里，迷宫有一个出口，Ignatius只有在炸弹爆炸前到达出口，才能出去。炸弹开始设定的时间是6分钟，为了避免炸弹因为震动而爆炸，Ignatius必须移动的很慢，只能走离他最近的前后左右（that is, if Ignatius stands on (x,y) now, he could only on (x+1,y), (x-1,y), (x,y+1), or (x,y-1) in the next minute)，每走一步花费1分钟。

其中有些地方是炸弹的重置区，可以把炸弹重置为6.

给你Ignatius的位置和迷宫让你判断他是否能走出迷宫：

（1）能 输出走出迷宫花费的最少时间;

（2）否 输出-1;

迷宫：

“2” 代表入口

“1” 可以走

“0” 是墙，不能走：

“4” 代表重置区：如果刚好走到4，准备reset引爆时间。但是走到该点时引爆时间  刚好变为0，则不能重新设置

“3” 代表出口    如果刚好走到3，炸弹引爆时间变为0，则算走不出迷宫。

   假设我们走的方向为 上 右 下 左

 

 

 

 

 

 

#include<iostream>
#include<string>
#include<string.h>
#include<sstream>
#include<queue>
using namespace std;
struct node 
{
	int x,y,step,time;
}s,e;
int n,m,a[15][15];
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
void bfs()
{
	queue<node>q;
	node t,x1;
	q.push(s);
	while(!q.empty())
	{
		t=q.front();
		q.pop();
		for(int i=0;i<4;i++)
		{
			x1.step=t.step+1;
			x1.time=t.time-1;
			x1.y=t.y+dir[i][1];
			x1.x=t.x+dir[i][0];
			if(x1.x>=0&&x1.x<n&&x1.y>=0&&x1.y<m&&a[x1.x][x1.y]!=0&&x1.time>0)
            {
                if(a[x1.x][x1.y]==3)
                {
                    cout<<x1.step<<endl;
                    return;
                }
                else if(a[x1.x][x1.y]==4)
                {
                  x1.time=6;
                  a[x1.x][x1.y]=0;
                }
                q.push(x1);
		}
	}
}
cout<<-1<<endl;
}//不要重复走4号位，二是可以走回头路
int main()
{
	int num;
	cin>>num;
	while(num--)
	{
		cin>>n>>m;
		for(int i=0;i<n;i++)
			for(int j=0;j<m;j++)
			{
				cin>>a[i][j];
				if(a[i][j]==2)
				{
					s.x=i;
					s.y=j;
					s.step=0;
					s.time=6;
				}
			}
		bfs();
	}
}