POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 87429 Accepted: 27431

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

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.

Source

USACO 2007 Open Silver

从一个数n到另一个数k，有三种操作，加1减1乘2，问最少几步能把n变成k

BFS水题，温暖一下。

#include <cstdio>#include <cstring>#include <queue>#define MAXN 100000+5using namespace std;int step[MAXN];int visited[MAXN];queue<int> nodeQueue;int bfs(int n,int k){    while(!nodeQueue.empty())        nodeQueue.pop();    step[n]=0;    visited[n]=1;    nodeQueue.push(n);    while(!nodeQueue.empty()){        int current=nodeQueue.front();        nodeQueue.pop();        int next;        for(int i=0;i<3;i++){            if(i==0)                next=current-1;            if(i==1)                next=current+1;            if(i==2)                next=current*2;            if(next<0||next>MAXN)                continue;            if(!visited[next]){                nodeQueue.push(next);                step[next]=step[current]+1;                visited[next]=1;            }            if(current==k)                return step[current];        }    }}int main() {    int n,k;    while(scanf("%d%d",&n,&k)>0){        memset(step,0, sizeof(step));        memset(visited,0, sizeof(visited));        if(n>=k)            printf("%d\n",n-k);        else            printf("%d\n",bfs(n,k));    }    return 0;}