http://poj.org/problem?id=3278(bfs)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 76935		Accepted: 24323

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.

 

题意：给出两个数star，end，给出x-1，x+1，x*2四种运算，使star变成end;

 

思路：直接bfs就好了；

 

代码：

 1 #include <iostream>
 2 #include <string.h>
 3 #include <queue>
 4 #define MAXN 200000+10
 5 using namespace std;
 6 
 7 struct Node       //***用结构体表示节点，x表示当前值，step记录当前步数
 8 {
 9     int x, step;
10 };
11 
12 int star, end, vis[MAXN];
13 
14 int bfs(int x)
15 {
16     Node a, b, c;
17     a.x = x, a.step = 0;      //***赋初值
18     queue<Node>q;            //***用队列实现
19     q.push(a);                //***头节点入队
20     while(!q.empty())         //***如果队列不为空，继续循环
21     {
22         b = q.front();        //***取当前队列的头节点做父节点并将其从队列中删除
23         q.pop();
24         vis[b.x] = 1;       //***标记
25         if(b.x == end)       //***如果达到目标，返回步数
26         {
27             return b.step;
28         }
29         c = b;                  //***符合条件的儿子入队
30         if(c.x >= 1 && !vis[c.x-1])
31         {
32             c.x-=1;
33             c.step++;
34             vis[c.x]=1;
35             q.push(c);
36         }
37         c = b;
38         if(c.x < end && !vis[c.x+1])
39         {
40             c.x+=1;
41             c.step++;
42             vis[c.x]=1;
43             q.push(c);
44         }
45         c = b;
46         if(c.x < end && !vis[2*c.x])
47         {
48             c.x*=2;
49             c.step++;
50             vis[c.x]=1;
51             q.push(c);
52         }
53     }
54     return -1;
55 }
56 
57 int main(void)
58 {
59     while(cin >> star >> end)
60     {
61         if(star >= end)             //***如果star>=end，则每次减1，最少需要star-end步
62         {
63             cout << star - end << endl;
64             continue;
65         }
66         memset(vis, 0, sizeof(vis));  //***标记数组清0，不然会对后面的测试产生影响
67         int ans=bfs(star);            //***深搜
68         cout << ans << endl;
69     }
70     return 0;
71 }

 

转载于:https://www.cnblogs.com/geloutingyu/p/5858192.html