POJ3278——Catch That Cow(BFS)

Catch That Cow

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：

　　　　有个农夫在N点，有个牛在K点。农夫有三种移动方式：1）向前一步 2）向后一步 3）从当所在位置X瞬间移动到2*X点

　　　　输出最少进行多少步，农夫能找到牛。(大坑是农夫和牛的活动范围在[0,100000])

解题思路：

　　　　bfs即可。注意农夫的可活动范围

ps:HUD 2717与这个题一样，但是是多组输入输出，请注意！

Code：

#include<string>
#include<cstring>
#include<cstdio>
#include<queue>
#include<iostream>
using namespace std;
int dis[100005],vis[100005],N,K;
queue<int> q;
int bfs(int N,int K)
{
    q.push(N);
    dis[N]=0,vis[N]=1;
    while (!q.empty())
    {
        int x[5],i;
        int front=q.front();
        q.pop();
        x[1]=front-1,x[2]=front+1,x[3]=front*2;
        for (i=1; i<=3; i++)
        {
            if (x[i]>=0&&x[i]<=100000&&!vis[x[i]])
            {
                q.push(x[i]),vis[x[i]]=1,dis[x[i]]=dis[front]+1;
                if (x[i]==K) return dis[x[i]];
            }
        }
    }
}
int main()
{
    cin>>N>>K;
    if (N==K) cout<<0;
    else
        cout<<bfs(N,K);
    return 0;
}