poj 3278 Catch That Cow_catch that cowcsdn-优快云博客

本文链接：https://blog.youkuaiyun.com/lpeaceminusone/article/details/78535365

本文介绍了一个经典的算法问题——如何在最短时间内通过行走或瞬移捕捉到静止不动的牛。文章提供了一段C++代码实现，利用广度优先搜索算法找到从起点到目标点的最短路径。

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Catch That Cow
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 100270 Accepted: 31366
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.

题意：从x走到y点，每一步都有三种方案，走到2倍的位子或者加1或者减1。

思路：找的是最短的走法，是广搜的模版题。

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<queue>
#include<string>
using namespace std;
struct node
{
    int x,l;
} nod[100010];
int n,k,ans;
int vis[200010];
int bfs()
{
    queue<node>q;
    while(!q.empty())
        q.pop();
    vis[n]=1;
    node t;
    t.x=n;
    t.l=0;
    q.push(t);
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        if(t.x==k)
            return t.l;

        for(int j=0; j<3; j++)
        {
            node T=t;
            if(j==0)
            {
                T.x*=2;
                T.l+=1;
            }
            if(j==1)
            {
                T.x+=1;
                T.l+=1;
            }
            if(j==2)
            {
                T.x-=1;
                T.l+=1;
            }
            if(T.x==k)
                return T.l;
            if(T.x<200005&&T.x>=0&&vis[T.x]==0)
            {
                vis[T.x]=1;
                q.push(T);
            }
        }
    }
    return 0;
}
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        memset(vis,0,sizeof(vis));
        cout<<bfs()<<endl;
    }
}