POJ3278Catch That Cow（BFS）

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 70652 Accepted: 22216

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX + 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 andK

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.

/**题意：有一个农民和一头牛，他们在一个数轴上，牛在k位置保持不动，农户开始时在n位置。*设农户当前在M位置，每次移动时有三种选择：*1.移动到M-1；*2.移动到M+1位置；*3.移动到M*2的位置。*问最少移动多少次可以移动到牛所在的位置。所以可以用广搜来搜索这三个状态，直到搜索到牛所在的位置。**/#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;const int N = 200100;int n, k;struct node{    int x, step;};queue<node> Q;int vis[N];void BFS(){   //搜索关于X的三种状态    int X, STEP;    while(!Q.empty())    {        node tmp = Q.front();        Q.pop();X = tmp.x;STEP = tmp.step;        if(X == k)        {            printf("%d\n",STEP);            return ;        }        if(X >= 1 && !vis[X - 1]) //要保证减1后有意义,所以要X >= 1        {            node temp;            vis[X - 1] = 1;            temp.x = X - 1;            temp.step = STEP + 1;            Q.push(temp);        }        if(X <= k && !vis[X + 1])        {            node temp;            vis[X + 1] = 1;            temp.x = X + 1;            temp.step = STEP + 1;            Q.push(temp);        }        if(X <= k && !vis[X * 2])        {            node temp;            vis[X * 2] = 1;            temp.x = 2 * X;            temp.step = STEP + 1;            Q.push(temp);        }    }}int main(){    while(~scanf("%d%d",&n,&k))    {        while(!Q.empty()) Q.pop();        memset(vis,0,sizeof(vis));        vis[n] = 1;        node t;        t.x = n, t.step = 0;        Q.push(t);BFS();    }    return 0;}