hdu 1548 A strange lift 宽搜bfs+优先队列

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1548

There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP" , you will go up Ki floor,i.e,you will go to the i+Ki th floor,as the same, if you press the button "DOWN" , you will go down Ki floor,i.e,you will go to the i-Ki th floor. Of course, the lift can't go up high than N,and can't go down lower than 1. For example, there is a buliding with 5 floors, and k1 = 3, k2 = 3,k3 = 1,k4 = 2, k5 = 5.Begining from the 1 st floor,you can press the button "UP", and you'll go up to the 4 th floor,and if you press the button "DOWN", the lift can't do it, because it can't go down to the -2 th floor,as you know ,the -2 th floor isn't exist.
Here comes the problem: when you are on floor A,and you want to go to floor B,how many times at least he has to press the button "UP" or "DOWN"?

 

题意描述：n层楼（1<=n<=200），每层楼一个数字Ki和两个按钮UP和DOWN，如果你在A层楼，A层楼的数字Ka=3，你可以按下UP按钮到A+Ka层楼，按下DOWN按钮到A-Ka层楼（前提是A+Ka和A-Ka都在n的范围里）。 现在你在A层楼，要到目的地B层楼，问按下按钮的最少次数。

算法分析：直接bfs暴搜即可，由于涉及到最少次数问题，我习惯性的用到了优先队列处理。

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<cstdlib>

 5 #include<cmath>

 6 #include<algorithm>

 7 #include<queue>

 8 #define inf 0x7fffffff

 9 using namespace std;

10 const int maxn=200+10;

11 const int M = 40000+10;

12 

13 int n,A,B;

14 int an[maxn];

15 struct node

16 {

17     int cnt,id;

18     int value;

19     friend bool operator < (node a,node b)

20     {

21         return a.cnt > b.cnt;

22     }

23 }cur,tail;

24 

25 int vis[maxn];

26 int bfs()

27 {

28     priority_queue<node> Q;

29     cur.id=A ;cur.cnt=0 ;

30     cur.value=an[A];

31     Q.push(cur);

32     memset(vis,0,sizeof(vis));

33     vis[A]=1;

34     int count=0;

35     while (!Q.empty())

36     {

37         cur=Q.top() ;Q.pop() ;

38         if (cur.id==B) return cur.cnt;

39 

40         //cout<<cur.id<<" "<<an[cur.id]<<" "<<cur.cnt<<endl;

41         tail.id=cur.id+an[cur.id];

42         if (tail.id>=1 && tail.id<=n && !vis[tail.id])

43         {

44             tail.value=an[tail.id];

45             tail.cnt=cur.cnt+1;

46             vis[tail.id]=1;

47             Q.push(tail);

48         }

49 

50         tail.id=cur.id-an[cur.id];

51         if (tail.id>=1 && tail.id<=n && !vis[tail.id])

52         {

53             tail.value=an[tail.id];

54             tail.cnt=cur.cnt+1;

55             vis[tail.id]=1;

56             Q.push(tail);

57         }

58     }

59     return -1;

60 }

61 

62 int main()

63 {

64     while (scanf("%d",&n)!=EOF && n)

65     {

66         scanf("%d%d",&A,&B);

67         for (int i=1 ;i<=n ;i++) scanf("%d",&an[i]);

68         printf("%d\n",bfs());

69     }

70     return 0;

71 }