poj 3278 Catch That Cow

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

4

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.
 
 
不解释 超简单BFS 思想，值得注意的是有可能输入的N和K相等，此时应该输出0
其他没什么问题了
 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <queue>
 5 using namespace std;
 6 struct node
 7 {
 8     int x,time;
 9 };
10 int N,K;
11 int flag[100005];
12 bool judge(node p)
13 {
14     if(p.x<0 || p.x>100000 || p.time>flag[p.x])
15     {
16         return false;
17     }
18     return true;
19 }
20 void bfs()
21 {
22     node p,temp;
23     int i;
24     flag[N]=0;
25     p.x=N;
26     p.time=0;
27     queue<node>q;
28     while(!q.empty())
29     {
30         q.pop();
31     }
32     q.push(p);
33     while(!q.empty())
34     {
35         p=q.front();
36         q.pop();
37         for(i=0;i<3;i++)
38         {
39             temp=p;
40             if(i==0)
41             {
42                 temp.x+=1;
43             }
44             if(i==1)
45             {
46                 temp.x-=1;
47             }
48             if(i==2)
49             {
50                 temp.x*=2;
51             }
52             temp.time=p.time+1;
53             if(!judge(temp))
54             {
55                 continue;
56             }
57             if(temp.x==K)
58             {
59                 printf("%d\n",temp.time);
60                 return;
61             }
62             flag[temp.x]=temp.time;
63             q.push(temp);
64         }
65     }
66 }
67 int main()
68 {
69     while(~scanf("%d%d",&N,&K))
70     {
71         if(N==K)
72         {
73             printf("0\n");
74             continue;
75         }
76         memset(flag,0x7F,sizeof(flag));
77         bfs();
78     }
79     return 0;
80 }
View Code
posted @ 2013-05-30 22:41  欧阳生朵  阅读(2119)  评论(0)  编辑  收藏  举报