HDU 1072 Nightmare （广搜）

题目链接

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

分析：

刚开始拿到这道题的时候，以为就是和原先一样的普通的广搜，标记走过的路不能够再走，但是发现这样行不通，后来才发现并不是所有走过的路都不能走了，只是有炸弹重装装置的那个店不能够再走了，只需要标记这个点已经走过就行了。

0：表示墙，不能够走

1：空地，可以从这走

2：起始点

3：终点

4：炸弹重装装置

题目要求算出从其实带你到终点所花费的时间，他每走一个点就会花费一分钟。但是他的身上还有个距离爆炸时间为6分钟的炸弹，必须的保证他在走到每个点的时候炸弹没有爆炸，而且当他走到炸弹重装装置的时候，炸弹的爆炸时间就会又变为6分钟。

代码：

    #include<iostream>
    #include<stdio.h>
    #include<queue>
    using namespace std;
    int map[8][8];
    int n,m;
    struct node
    {
     int x,y,step,time;  //step为走的步数，time为bomb离爆炸时间
    }start;
    void store_map()//将地图的形式表示出来，并且找到起始点
    {
     for(int i=0;i<n;i++)
     {
      for(int j=0;j<m;j++)
      {
       scanf("%d",&map[i][j]);
       if(map[i][j]==2)
       {
        start.x=i;
        start.y=j;
        start.step=0;
        start.time=6;
       }
      }
     }
    }
    void bfs()
    {
    const int help[4][2]={{1,0},{-1,0},{0,1},{0,-1}};//能走的四个方向
     queue<node> q;
     q.push(start);
        node p1,p2;
     int i;
     while(!q.empty())
     {
      p1=q.front();
      q.pop();
      for(i=0;i<4;i++)
      {
       p2.step=p1.step+1;//步数加
       p2.time=p1.time-1;//时间减
       p2.x=p1.x+help[i][0];
       p2.y=p1.y+help[i][1];
       if(p2.x>=0 && p2.x<n && p2.y>=0 && p2.y<m && map[p2.x][p2.y]!=0 && p2.time>0)
       //注意这里的p2.time>0，意味着炸弹没有爆炸
       {
        if(map[p2.x][p2.y]==3)//重点的话就输出
        {
         printf("%d\n",p2.step);
         return;
        }
        else if(map[p2.x][p2.y]==4)//炸弹重装装置就把时间变为6
        {
         p2.time=6;
         map[p2.x][p2.y]=0;    //并且该位置4不能再访问了
        }
        q.push(p2);
       }
      }
     }
     printf("-1\n");//队空的话，也就意味着没有找到终点
    }
    int main()
    {
     int N;
     scanf("%d",&N);
     while(N--)
     {
      scanf("%d%d",&n,&m);
      store_map();
      bfs();
     }
     return 0;
    }