Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 45648 | Accepted: 14310 |
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.
Source
简单一维广搜。
题意：
你在一个一维坐标的n位置，牛在k位置，你要从n到k，抓到那头牛。你可以有三种走法，n+1，n-1，或者n*2直接跳。求你抓到那头牛的最短步数。
思路：
简单广搜的思想。状态跳转的时候有三种跳转的方式，将新的状态放到队列中，再不断提取队列中最前面的状态，直到找到k位置。
代码：
1 #include <iostream>
2 #include <stdio.h>
3 #include <string.h>
4 #include <queue>
5 using namespace std;
6
7 bool isw[100010];
8
9 struct Node{
10 int x;
11 int s;
12 };
13
14 bool judge(int x)
15 {
16 if(x<0 || x>100000)
17 return true;
18 if(isw[x])
19 return true;
20 return false;
21 }
22
23 int bfs(int sta,int end)
24 {
25 queue <Node> q;
26 Node cur,next;
27 cur.x = sta;
28 cur.s = 0;
29 isw[cur.x] = true;
30 q.push(cur);
31 while(!q.empty()){
32 cur = q.front();
33 q.pop();
34 if(cur.x==end)
35 return cur.s;
36 //前后一个个走
37 int nx;
38 nx = cur.x+1;
39 if(!judge(nx)){
40 next.x = nx;
41 next.s = cur.s + 1;
42 isw[next.x] = true;
43 q.push(next);
44 }
45 nx = cur.x-1;
46 if(!judge(nx)){
47 next.x = nx;
48 next.s = cur.s + 1;
49 isw[next.x] = true;
50 q.push(next);
51 }
52 //向前跳
53 nx = cur.x*2;
54 if(!judge(nx)){
55 next.x = nx;
56 next.s = cur.s + 1;
57 isw[next.x] = true;
58 q.push(next);
59 }
60 }
61 return 0;
62 }
63
64
65 int main()
66 {
67 int n,k;
68 while(scanf("%d%d",&n,&k)!=EOF){
69 memset(isw,0,sizeof(isw));
70 int step = bfs(n,k);
71 printf("%d
",step);
72 }
73 return 0;
74 }
Freecode : www.cnblogs.com/yym2013