HDU 2717&&poj 3278 Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13030    Accepted Submission(s): 4012

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
代码如下：
#include<stdio.h>#include<string.h>#include<math.h>#include<queue>#include<iostream>using namespace std;int con[200010]; //防止越界，考虑最坏情况2*x int n,k;typedef struct node{int x,t;}N;queue<N>q;void start(){while(!q.empty())q.pop();memset(con,0,sizeof(con));}int bfs(int n){N temp;temp.x=n;temp.t=0;con[n]=1;q.push(temp);while(!q.empty()){temp=q.front();q.pop();if(temp.x==k)return temp.t;else{N m;m.x=temp.x+1;m.t=temp.t+1;//不越界且未访问 if(!con[m.x] && m.x>=0 && m.x<=100000){q.push(m);con[m.x]=1;}m.x=temp.x-1;if(!con[m.x] && m.x>=0 && m.x<=100000){q.push(m);con[m.x]=1;}m.x=temp.x*2;if(!con[m.x] && m.x>=0 && m.x<=100000){q.push(m);con[m.x]=1;}}}}int main(){while(~scanf("%d%d",&n,&k)){start();printf("%d\n",bfs(n));}return 0;}