HDU 2717 Catch That Cow （深搜）

题目链接

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



HintThe 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.

分析：

每次走的话可以有三种走法，-1,+1,*2

代码：

#include<iostream>
#include<stdio.h>
#include<string>
#include<cstring>
#include<math.h>
#include<queue>
using namespace std;
int N=1000000;
int a[1010000]={0};防止走重，即第一步走过了后面不应该再走一次
struct ft//结构体，包含数和步数
{
    int x,y;
}p;
int k;
int bfs(int n)
{
    queue<ft>Q;//建队
    p.x=n,p.y=0;
    a[n]=1;
    Q.push(p);入队
    ft cur, nex;
    while(!Q.empty())
    {
        cur=Q.front();取队首元素
        Q.pop();出队
        if(cur.x==k)return cur.y;//判断是否达到目的
        nex.x=cur.x+1;//判断+1的操作是否合法
        if(nex.x>=0&&nex.x<N&&a[nex.x]==0)
        {
            nex.y=cur.y+1;
            a[nex.x]=1;
            Q.push(nex);合法则入队并步数+1
        }
        nex.x=cur.x-1;
        if(nex.x>=0&&nex.x<N&&a[nex.x]==0)
        {
            nex.y=cur.y+1;
            a[nex.x]=1;
            Q.push(nex);
        }
        nex.x=cur.x*2;
        if(nex.x>=0&&nex.x<N&&a[nex.x]==0)
        {
            nex.y=cur.y+1;
            a[nex.x]=1;
            Q.push(nex);
        }
    }
    return -1;
}
int main()
{
    int n;
    while(~scanf("%d %d",&n,&k))
    {
        memset(a,0,sizeof(a));//多次运行，刷0；
        int bs=bfs(n);
        if(bs>=0)
        printf("%d\n",bs);
    }
    return 0;
}