题目链接：https://vjudge.net/problem/POJ-3278

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.

题目大意： 花花经过长时间的研究，终于研发出了能够跃迁的宇宙飞船。现在，他想要前往致远星。假设地球和致远星都在一个坐标轴上，其中地球位于坐标n，而致远星位于坐标k。而花花的飞船支持以下两种运动方式：

飞行：在一个时间单位中，能够从坐标x移动到x-1或x+1；

跃迁：在一个时间单位中，能够直接从x跃迁到2x。

现在，花花想知道，他需要多长时间才能到达致远星？

解题思路：利用BFS求出n到k的最短距离，关键在于对搜索过程进行剪枝，省去重复的工作。

Cpp Code

#include

#include

#include

using namespace std;

const int maxn = 1e5 + 5;

int n, k;

bool vis[maxn];//标记数组

int ans[maxn];

queue q;

int bfs(int x,int k){

int cur, next;

ans[x] = 0;

vis[x] = 1;//访问后标记为1，防止重复访问

q.push(x);

while (!q.empty())

{

cur = q.front();

q.pop();

for (int i = 0; i < 3;i++){//三次转换

if(i==0){

next = cur - 1;

}

else if(i==1){

next = cur + 1;

}

else{

next = cur*2;

}

//符合条件才压入队列

if(next>=0&&next<=100000&&vis[next]==0){

vis[next] = 1;

q.push(next);

ans[next] = ans[cur] + 1;

}

if(next==k){

return ans[next];

}

}

/* code */

}

}

int main(){

cin >> n >> k;

fill(vis, vis + maxn, 0);

while (!q.empty())

{

q.pop();

}

if(n>=k){

cout << n - k << endl;

}

else{

cout << bfs(n,k) << endl;

}

return 0;

}

Java Code

import java.util.LinkedList;

import java.util.Queue;

import java.util.Scanner;

public class Main {

static class node{

int x;//数的值

int num;//次数

node(){}

node(int a,int b){

x=a;

num=b;

}

}

static boolean[] vis=new boolean[(int)1e5+5];//标记数组

static boolean ok(int x) {//判断函数

if(x>=0&&x<=(int)1e5&&vis[x]==false) {

return true;

}

return false;

}

public static int BFS(int n,int k) {

Queue q=new LinkedList();

vis[n]=true;//访问后标记

q.offer(new node(n,0));

node op;

while(!q.isEmpty()) {

op=q.poll();

if(op.x==k) {

return op.num;

}

int x,num=op.num+1;

for(int i=0;i<3;i++) {//三次转换

if(i==0) {

x=op.x-1;

}

else if(i==1) {

x=op.x+1;

}

else {

x=op.x*2;

}

if(ok(x)) {

vis[x]=true;//访问后标记

q.offer(new node(x,num));

}

}

}

return -1;

}

public static void main(String[] args) {

Scanner sc=new Scanner(System.in);

int n,k;

n=sc.nextInt();

k=sc.nextInt();

System.out.print(BFS(n,k));

// TODO Auto-generated method stub

}

}