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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
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