POJ

Catch That Cow

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.

 

 

开始考虑用dfs做。。没考虑双向问题，向左向右会陷入死循环。还是BFS大法好，，求最短路径问题，遍历部分点即可。

 

 

#include<stdio.h>
#include<queue>
using namespace std;

struct Node{
    int x,y;
}node;
int b[200010];

int main()
{
    int n,k;
    queue<Node> q;
    scanf("%d%d",&n,&k);
    if(n==k) printf("0
");    //特判
    else{
    node.x=n;
    node.y=0;
    q.push(node);
    b[n]=1;
    while(q.size()){
        node.x=q.front().x*2;
        node.y=q.front().y+1;
        if(node.x<=200010&&node.x>=0){
            if(node.x==k){
                printf("%d
",node.y);
                break;
            }
            if(b[node.x]==0){
                b[node.x]=1;
                q.push(node);
            }
        }
        node.x=q.front().x+1;
        node.y=q.front().y+1;
        if(node.x<=200010&&node.x>=0){
            if(node.x==k){
                printf("%d
",node.y);
                break;
            }
            if(b[node.x]==0){
                b[node.x]=1;                
                q.push(node);
            }
        }
        node.x=q.front().x-1;
        node.y=q.front().y+1;
        if(node.x<=200010&&node.x>=0){
            if(node.x==k){
                printf("%d
",node.y);
                break;
            }
            if(b[node.x]==0){
                b[node.x]=1;                
                q.push(node);
            }
        }
        q.pop();
    }
    }
    return 0;
}