HDU-2717 Catch That Cow（搜索）

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16648 Accepted Submission(s): 4959

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.

没什么 好说的简单搜索 ， 记住可以后退一步就行 ，卡了一下 想想挺傻逼的

code：

#include<bits/stdc++.h>using namespace std;int vis[100100];struct node{    int x;    int step;};node s1,s2;int bfs(int n,int k){    queue<node>q;    s1.x=n;    s1.step=0;    memset(vis,0,sizeof(vis));    q.push(s1);    while(!q.empty())    {        s1=q.front();        q.pop();        if(s1.x==k)            return s1.step;        s2.x=s1.x+1;        s2.step=s1.step+1;        if(s2.x>=0&&s2.x<=100000&&!vis[s2.x])        {            vis[s2.x]=1;            q.push(s2);        }        s2.x=s1.x-1;        s2.step=s1.step+1;        if(s2.x>=0&&s2.x<=100000&&!vis[s2.x])        {            vis[s2.x]=1;            q.push(s2);        }        s2.x=2*s1.x;        s2.step=s1.step+1;        if(s2.x>=0&&s2.x<=100000&&!vis[s2.x])        {            vis[s2.x]=1;            q.push(s2);        }    }}int main(){    int n,k;    while (~scanf("%d%d",&n,&k))    {        printf ("%d\n",bfs(n,k));    }    return 0;}