POJ-3278 Catch That Cow (BFS入门题

Catch That Cow

Time Limit: 2000MS

Memory Limit: 65536K

Total Submissions: 93780

Accepted: 29419

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 pointsX - 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

分析：bfs入门题如果前面的数比后面的数大，那么只能后退，直接输出两者之差。一般情况下，我们用搜索直接解决，每一步三个操作，搜出答案即可。注意点见注释。

#include <iostream>#include<stdio.h>#include<queue>#include<string.h>using namespace std;const int SIZE=100005;int step[3*SIZE]; //防止数组越界的保险，正常没有3也可以int check[3*SIZE];void bfs(int s,int e){    queue<int> q;    while(!q.empty()) q.pop();    q.push(s);    step[s]=0;    check[s]=1;    while(!q.empty())    {        int now=q.front();        if(now==e) break;        q.pop();        if(!check[now+1]&&now+1<=100000)        {            q.push(now+1);            step[now+1]=step[now]+1;            check[now+1]=1; //这个地方我觉得是BFS最需要注意的，如果写在最上方的话，后面就会有一些情况第二次进入队列在步数上第二次计算就会有问题。        }        if(!check[now-1]&&now-1>=0)        {            q.push(now-1);            step[now-1]=step[now]+1;            check[now-1]=1;        }        if(!check[now*2]&&now*2<=100000)        {            q.push(now*2);            step[now*2]=step[now]+1;            check[now*2]=1;        }    }}int main(){    int m,n;    while(cin>>m>>n)    {        if(m>=n)        {            cout<<m-n<<endl;            continue;        }        memset(step,0,sizeof(step));        memset(check,0,sizeof(check));        bfs(m,n);        cout<<step[n]<<endl;    }    return 0;}