bzoj 1646 [Usaco2007 Open]【Catch That Cow 抓住那只牛】

描述

　　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?

　　农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来,他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有两种办法移动，步行和瞬移：步行每秒种可以让约翰从x处走到x+1或x-1处；而瞬移则可让他在1秒内从x处消失，在2x处出现．然而那只逃逸的奶牛，悲剧地没有发现自己的处境多么糟糕，正站在那儿一动不动．那么，约翰需要多少时间抓住那只牛呢？

输入输出格式

输入

　　* Line 1: Two space-separated integers: N and K
　　仅有两个整数N和K.

输出

　　* Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
　　最短的时间．

输入输出样例

输入样例1

5 17

输出样例1

4

解题思路

　　这里特判n>=k的情况，然后搜索三种方式，再剪个枝就行了。

题解

#include<bits/stdc++.h>
using namespace std;
int n,k;
queue<int> q; 
int flag[1000001];//标记 
void bfs()
{
    q.push(n);
    flag[n]=1;
    while(!q.empty())
    {
        int head=q.front();
        q.pop();
        if(head==k)break;
        if(!flag[head+1]&&head<2*k)//瞬移（如果head>=2*k就绝对浪费了步数，不是最优的） 
        {
            flag[head+1]=flag[head]+1;
            q.push(head+1);
        }
        if(!flag[head-1]&&head<2*k&&head>1)//向后走 （没有负数） 
        {
            flag[head-1]=flag[head]+1;
            q.push(head-1);
        }
        if(!flag[head*2]&&head<2*k)//向前走 
        {
            flag[head*2]=flag[head]+1;
            q.push(head*2);
        }
    }
    cout<<flag[k]-1;
} 
int main()
{
    cin>>n>>k;
    if(n>=k)//特判，只能一步一步向回走 
    {
        cout<<n-k;
        return 0;
    }
    bfs();
    return 0;
}