poj 3278 BFS

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 30928		Accepted: 9537

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

一般的广搜，不过要考虑的就是不要让数组越界

代码：

View Code

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 using namespace std;
 5 
 6 struct M{
 7     int add;
 8     int step;
 9 };
10 
11 M queue[2000005];
12 bool vis[2000005]={false};
13 
14 int BFS(int s, int e)
15 {
16     
17     int head=0;
18     int tail=0;
19     queue[0].add=s;
20     queue[tail++].step=0;
21     while(head<tail)
22     {
23         M x=queue[head++];
24         if(x.add==e)
25         {
26             return x.step;
27         }
28         if(!vis[x.add+1])
29         {
30             vis[x.add+1]=true;
31             queue[tail].add=x.add+1;
32             queue[tail++].step=x.step+1;
33         }
34         if( x.add-1>=0 && !vis[x.add-1])    //x.add-1>=0 这个不要忘了，不然会runtime error
35         {
36             vis[x.add-1]=true;
37             queue[tail].add=x.add-1;
38             queue[tail++].step=x.step+1;
39         }
40         if(!vis[x.add*2] && x.add*2<=200000)  // 同样的，add*2<=200000也是必须的，否则由于广搜，会乘到很大。
41         {
42             vis[x.add*2]=true;
43             queue[tail].add=x.add*2;
44             queue[tail++].step=x.step+1;
45         }
46     }
47     return 0;
48 }
49 
50 int main()
51 {
52     int N,K;
53     int i,j;
54     while(scanf("%d%d",&N,&K)!=EOF)
55     {
56         printf("%d\n",BFS(N,K));
57     }
58     return 0;
59 }