HDUOJ-----(1072)Nightmare(bfs)

Nightmare

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5795    Accepted Submission(s): 2868

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

bfs 的应用范围：

求最值问题.....属于盲目搜索的一种....

特点是所用的内存大相对于dfs....

代码如下：

 1 #include<cstdio>

 2 #include<iostream>

 3 #include<deque>

 4 #define maxn 10

 5 #define SET 6

 6 using namespace std;

 7 

 8 int map[maxn][maxn],nn,mm;

 9 int dir[4][2]=

10 {

11     {0,1},  /*向右*/

12     {0,-1}, /*向左*/

13     {-1,0},  /*向下*/

14     {1,0}  /*向上*/

15 } ;

16 struct node{

17  int x,y;         /*记录位置*/

18  int ans,time;   /*步数和时间*/

19 }start;

20  

21 /*生成地图*/

22 void save_map()

23 {

24     for(int i=0;i<nn;i++)

25     {

26         for(int j=0;j<mm;j++)

27         {

28             scanf("%d",&map[i][j]);

29             if(map[i][j]==2)

30             {

31                 start.x=i;

32                 start.y=j;

33                 start.ans=0;

34                 start.time=SET;  /*设定为6s*/

35             }

36         }

37     }

38   return ;

39 }

40 

41 void bfs( void )

42 {

43     deque<node>q ;    /*队列实现*/ 

44     

45     node q1,q2;    //暂存

46     q.push_back(start);      //将start的位置初始化.....

47     

48     /* 当队列不为空的时候，执行下列程序 */

49     while(!q.empty())

50     {

51       q1=q.front();     //将对头的数据拿出来--->q1;

52       q.pop_front();     //q.pop()一样

53       for(int i=0; i<4 ;i++)

54       {

55         /*进入下一个状态*/

56         q2.x=q1.x+dir[i][0];  

57         q2.y=q1.y+dir[i][1];

58         q2.ans=q1.ans+1;

59         q2.time=q1.time-1;

60        //如果满足这些条件便可进行下一步搜索    

61        if(q2.x>=0&&q2.y>=0&&q2.x<nn&&q2.y<=mm&&map[q2.x][q2.y]!=0&&q2.time>0)

62        {

63          /*说明找到了答案，可以结束了*/

64          if(map[q2.x][q2.y]==3)

65          {

66            printf("%d\n",q2.ans);

67 

68            return ;

69          }

70          else if(map[q2.x][q2.y]==4)

71          {

72             /*重置时间，其他照常*/

73             q2.time=SET;

74             map[q2.x][q2.y]=0; //走过，不可再走，不然没完没了

75          }

76           q.push_back(q2);

77        }

78       }

79     }

80     /*说明不存在*/

81     printf("-1\n");

82     return ;

83 }

84     

85 int main()

86 {

87     int test;

88     scanf("%d",&test);

89     while(test--)

90     {

91      scanf("%d %d",&nn,&mm);

92      save_map();

93      bfs();

94     }

95     return 0;

96 }

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