poj 3278:Catch That Cow（简单一维广搜）

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 }

 

Freecode : www.cnblogs.com/yym2013