Catch That Cow（广搜）

个人心得：其实有关搜素或者地图啥的都可以用广搜，但要注意标志物不然会变得很复杂，想这题，忘记了标志，结果内存超时；

将每个动作扔入队列，但要注意如何更简便，更节省时间，空间

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?

InputLine 1: Two space-separated integers: N and KOutputLine 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.
 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<queue>
 5 using namespace std;
 6 int sum;
 7 int ok=0;
 8 struct Node
 9 {
10     int x;
11     int y;
12 
13 };
14 int book[100005];
15 
16 void dfs(int n,int m)
17 {
18     memset(book,0,sizeof(book));
19     queue<Node >s;
20     book[n]=1;
21     Node t;
22     t.x=n;t.y=0;
23     s.push(t);
24     int a;
25     Node tt;
26     while(!s.empty())
27     {
28           a=s.front().x*2;
29         tt.x=a,tt.y=s.front().y+1;
30         if(a==m)
31         {
32             sum=tt.y;
33             return ;
34 
35         }
36          if(tt.x>=0&&tt.x<=100000)
37             if(!book[a])
38            {
39                book[a]=1;
40                s.push(tt);
41 
42                }
43         a=s.front().x+1;
44         tt.x=a;
45         tt.y=s.front().y+1;
46         if(a==m)
47         {
48             sum=tt.y;
49            return ;
50 
51         }
52       if(tt.x>=0&&tt.x<=100000)
53             if(!book[a])
54            {
55                book[a]=1;
56                s.push(tt);
57 
58                };
59          a=s.front().x-1;
60         tt.x=a,tt.y=s.front().y+1;
61         if(a==m)
62         {
63             sum=tt.y;
64             return ;
65 
66         }
67         if(tt.x>=0&&tt.x<=100000)
68             if(!book[a])
69            {
70                book[a]=1;
71                s.push(tt);
72 
73                }
74         s.pop();
75 
76     }
77     return ;
78 
79 
80 }
81 int main()
82 {
83 
84     int n,m;
85     while(cin>>n>>m)
86     {
87         sum=0;
88         if(n>=m) sum=n-m;
89        else
90         dfs(n,m);
91     cout<<sum<<endl;
92 
93     }
94     return 0;
95 
96 }

 

posted @ 2017-07-25 21:01  余生漫漫浪  阅读(243)  评论(0)  编辑  收藏  举报