hdu 1548 A strange lift (BFS)

A strange lift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9424    Accepted Submission(s): 3562


Problem Description
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"?
 

 

Input
The input consists of several test cases.,Each test case contains two lines.
The first line contains three integers N ,A,B( 1 <= N,A,B <= 200) which describe above,The second line consist N integers k1,k2,....kn.
A single 0 indicate the end of the input.
 

 

Output
For each case of the input output a interger, the least times you have to press the button when you on floor A,and you want to go to floor B.If you can't reach floor B,printf "-1".
 

 

Sample Input
5 1 5 3 3 1 2 5 0
 

 

Sample Output
3
 

 

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 1 //0MS    276K    1053 B    C++
 2 /*
 3 
 4     题意：
 5         有n层楼，每层有有一个值ki，表示在第i层可以上ki层或下ki层，问是否可以从第a层到第b层，
 6     可以则输出最少步数，不可以输出-1
 7     
 8     BFS：
 9         典型的BFS，一段时间没写，代码有点挫.. 
10 
11 */
12 #include<iostream>
13 #include<queue>
14 #define N 205
15 using namespace std;
16 struct node{
17     int id,cnt;
18 };
19 int st[N];
20 int vis[N];
21 int n,s,e;
22 int bfs()
23 {
24     node t={s,0};
25     memset(vis,0,sizeof(vis));
26     vis[s]=1;
27     queue<node>Q;
28     Q.push(t);
29     while(!Q.empty()){
30         t=Q.front();
31         Q.pop();
32         node tt=t;
33         if(t.id==e) return t.cnt;
34         tt.cnt++;
35         if(tt.id+st[t.id]>=1 && tt.id+st[t.id]<=n){
36             tt.id+=st[t.id];
37             if(!vis[tt.id]){
38                 vis[tt.id]=1;
39                 Q.push(tt);
40             }
41         }
42         tt.id=t.id;
43         if(tt.id-st[t.id]>=1 && tt.id-st[t.id]<=n){
44             tt.id-=st[t.id];
45             if(!vis[tt.id]){
46                 vis[tt.id]=1;
47                 Q.push(tt);
48             }
49         }
50     }
51     return -1;   
52 } 
53 int main(void)
54 {
55     while(scanf("%d",&n),n)
56     {
57         scanf("%d%d",&s,&e);
58         for(int i=1;i<=n;i++)
59             scanf("%d",&st[i]);
60         printf("%d\n",bfs());
61     }
62     return 0;
63 }

 

posted @ 2013-12-10 11:37  heaventouch  阅读(211)  评论(0)  编辑  收藏  举报