Catch That Cow(剪枝防止超时)

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

*由题意得，此题直接反应的是一颗度为三的树。遍历此树可得步数，但是如果不对树进行处理会超时；其他问题都很好想。

#include
#include
#include
using namespace std;
typedef struct node{
    int data;
    int ceng;
}Tree;
bool vis[100000];
queue  q;
int n,k;
void bfs(){
    Tree n;
    while(q.front().data!=k){
        int a=q.front().data;
        int b=q.front().ceng;
        n.data=a+1;
        n.ceng=b+1;
        if(n.data>=0&&n.data<=100000&&vis[n.data]==false){//这就是对数据的处理，将超出题目范围的节点和访问过得节点筛出。
        q.push(n);                                           //将结构体节点插入队列
        vis[n.data]=true;
        }
        n.data=a-1;
        n.ceng=b+1;
        if(n.data>=0&&n.data<=100000&&vis[n.data]==false){
        q.push(n);
        vis[n.data]=true;
        }
        n.data=a*2;
        n.ceng=b+1;
        if(n.data>=0&&n.data<=100000&&vis[n.data]==false){
        q.push(n);
        vis[n.data]=true;
        }
        q.pop();
    }
    cout<>n>>k;
    Tree t;
    memset(vis,false,sizeof(vis));
    t.ceng=0;
    t.data=n;
    q.push(t);
    bfs();
    return 0;
}