POJ 3278： Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 44613 Accepted: 13946

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

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BFS来搜索这三个状态，直到搜索到牛所在的位置。

#include<cstdio>#include<iostream>#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 vist[N];void bfs(){      int cow, ans;    while(!q.empty())    {        node tmp = q.front();        q.pop();        cow = tmp.x;        ans = tmp.step;        if(cow == k)        {            printf("%d\n",ans);            return ;        }        if(cow >= 1 && !vist[cow - 1]) //要保证减1后有意义,所以要cow >= 1    减一的情况        {            node temp;            vist[cow - 1] = 1;            temp.x = cow - 1;            temp.step = ans + 1;            q.push(temp);        }        if(cow <= k && !vist[cow + 1]) //加1的情况        {            node temp;            vist[cow + 1] = 1;            temp.x = cow + 1;            temp.step = ans + 1;            q.push(temp);        }        if(cow <= k && !vist[cow * 2]) //乘二的情况        {            node temp;            vist[cow * 2] = 1;            temp.x = 2 * cow;            temp.step = ans + 1;            q.push(temp);        }    }}int main(){    while(~scanf("%d%d",&n,&k))    {        while(!q.empty()) q.pop();        memset(vist,0,sizeof(vist));        vist[n] = 1;        node t;        t.x = n, t.step = 0;        q.push(t);        bfs();    }    return 0;}