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poj 3278:Catch That Cow（简单一维广搜）

2024-07-13 09:29:17   225人阅读

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

USACO 2007 Open Silver
 
　　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\n",step);72     }73     return 0;74 }

 

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