Catch That Cow POJ - 3278

题目链接:点我

题目

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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.

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题意:

你每次可以达到当前点的左边或者右边或者当前点2倍坐标的位置,问是否能达到目标.

思路:

简单的bfs,同时标记你走过的位置就可以了.

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

struct ss
{
    int x,step;
}q[100000+10];
bool vis[100000+10];
int n,k;

void bfs()
{
    memset(vis,false,sizeof(vis));
    int head=0;
    int tail=0;
    q[tail].x=n;
    q[tail++].step=0;
    vis[n]=true;
    while(head<tail){
        ss p=q[head++];
        if(p.x==k){
            cout<<p.step<<endl;
            return ;
        }
        if(p.x-1>=0&&vis[p.x-1]==false){
            q[tail].x=p.x-1;
            q[tail++].step=p.step+1;
            vis[p.x-1]=true;
        } if(p.x+1<=100000&&vis[p.x+1]==false){
            q[tail].x=p.x+1;
            q[tail++].step=p.step+1;
            vis[p.x+1]=true;
        } if(p.x*2 <= 100000&&vis[p.x*2]==false){
            q[tail].x=p.x*2;
            q[tail++].step=p.step+1;
            vis[p.x*2]=true;
        }
    }
}

int main()
{
    while(scanf("%d %d",&n,&k)!=EOF){
        bfs();
    }
    return 0;
}