杭电2717Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9932    Accepted Submission(s): 3109

Problem 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

 

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广搜，感觉自己变傻逼了，这都不会了。

#include<stdio.h>#include<queue>#include<string.h>#include<iostream>#include<algorithm>using namespace std;struct node{int start;int time;}a,team;int vist[110000],i,j,k,l,m,n,end,best;void bfs(int num){queue<node > q;a.start=num;a.time =0;q.push(a);memset(vist,0,sizeof(vist));vist[a.start]=1;while(!q.empty()){a=q.front();q.pop();for(i=0;i<3;i++){if(i==0)team.start =a.start +1;if(i==1)team.start =a.start -1;if(i==2)team.start =a.start *2;team.time =a.time +1;if(team.start ==n){if(best>=team.time)best=team.time;}if(team.start>100000||team.start <0)continue;if(!vist[team.start]){vist[team.start]=1;q.push(team);}}}}int main(){while(scanf("%d%d",&m,&n)!=EOF){best=99999;if(n<=m){printf("%d\n",m-n);continue;}bfs(m);printf("%d\n",best);}return 0; }