Catch That Cow
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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.
Analysis
从一个点到另一个点，每一步有多种走法，这是典型的 bfs 问题。
Tips
- 注意数组开的够不够大
- 在每一次开始判断是否已经搜到解
1 #include <stdio.h>
2 #include <string.h>
3 #include<queue>
4
5 using namespace std;
6
7 int vis[120000];
8 int n,k,flag;
9
10 struct Node
11 {
12 int x;
13 int tm;
14 }root;
15
16 void bfs()
17 {
18 memset(vis,0,sizeof(vis));
19 queue<Node>que;
20 que.push(root);
21
22 vis[root.x] = 1;
23 int ans=100005;
24
25 while(!que.empty())
26 {
27 Node q = que.front();
28 que.pop();
29
30 //printf("q.x = %d\n",q.x);
31
32 if(q.x == k )
33 {
34 ans = q.tm;
35 flag = 1;
36 break;
37 }
38
39 Node q1,q2,q3;
40
41 if(q.x+1 < 100005 &&!vis[q.x + 1] )
42 {
43 //printf("q2.x = %d\n",q.x+1);
44 q2.x = q.x + 1;
45 q2.tm = q.tm + 1;
46
47 vis[q2.x] = 1;
48 que.push(q2);
49 }
50
51 if(q.x-1 >=0 && !vis[q.x - 1] )
52 {
53 //printf("q1.x = %d\n",q.x-1);
54 q1.x = q.x - 1;
55 q1.tm = q.tm + 1;
56
57 vis[q1.x] = 1;
58 que.push(q1);
59 }
60
61
62
63 if(q.x *2 <100005 && !vis[q.x *2] )
64 {
65 //printf("q3.x = %d\n",q.x*2);
66 q3.x = q.x *2;
67 q3.tm = q.tm + 1;
68
69 vis[q3.x] = 1;
70 que.push(q3);
71
72 }
73 }
74 if(flag) printf("%d\n",ans);
75 //else printf("-1\n");
76 }
77
78
79 int main()
80 {
81
82 while(~scanf("%d%d",&n,&k))
83 {
84 flag = 0;
85
86 root.x = n;
87 root.tm = 0;
88
89 bfs();
90 }
91
92 return 0;
93 }
时间： 2024-12-11 13:07:06