Catch That Cow

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原文链接：http://www.cnblogs.com/wuwangchuxin0924/p/6036556.html

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#java

本文详细解析了USACO2007OpenSilver竞赛题目“CatchThatCow”的算法实现过程，该题要求找出农夫约翰最快捕捉逃跑奶牛的方法。通过广度优先搜索算法（BFS），文章给出了从起始位置到目标位置的最短路径解决方案。

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 243 Accepted Submission(s): 88

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

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

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#include<bits/stdc++.h>
#define N 5000100
#define INF 0x3f3f3f3f
using namespace std;
int n,m;
struct node
{
    int x,val;
    node(){}
    node(int a,int b)
    {
        x=a;
        val=b;
    }
};
bool vis[N];
int bfs()
{
    queue<node>q;
    q.push(node(n,0));
    vis[n]=true;
    while(!q.empty())
    {
        node fr=q.front();
        q.pop();
        //cout<<fr.x<<endl;
        if(fr.x==m)
        {
            //cout<<"x="<<fr.x<<endl;
            return fr.val;
        }
        int ans=fr.x+1;
        if(ans>=0&&ans<=100000&&!vis[ans])
        {
            q.push(node(ans,fr.val+1));
            vis[ans]=true;
        }
        ans=fr.x-1;
        if(ans>=0&&ans<=100000&&!vis[ans])
        {
            q.push(node(ans,fr.val+1));
            vis[ans]=true;
        }
        ans=fr.x*2;
        if(ans>=0&&ans<=100000&&!vis[ans])
        {
            q.push(node(ans,fr.val+1));
            vis[ans]=true;
        }
    }
    //return -1;
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(vis,false,sizeof vis);
        printf("%d\n",bfs());
    }
    return 0;
}