Catch That Cow(bfs)

最新推荐文章于 2022-07-29 10:49:56 发布

fanlinqiang

最新推荐文章于 2022-07-29 10:49:56 发布

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分类专栏：搜索

本文链接：https://blog.youkuaiyun.com/flqbestboy/article/details/8209882

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本文介绍了一个简单的搜索算法案例——“捉住那头牛”。该问题描述了农夫如何通过步行和传送的方式，在最短时间内找到静止不动的逃逸奶牛。文章提供了一段使用C++编写的广度优先搜索(BFS)算法实现。

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Catch That Cow

http://poj.org/problem?id=3278

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 32993		Accepted: 10140

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

我晕、sb了。。。一直没写边界条件wa^n

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int M=1000000;
struct node
{
    int x,t;
}n1,n2;
queue<node> Q;
bool mark[M];
int main()
{
    int a,b,res;
    while(cin>>a>>b)
    {
        memset(mark,true,sizeof(mark));
        while(!Q.empty())
         Q.pop();
        n1.x=a;n1.t=0;
        Q.push(n1);
        mark[a]=false;
        while(!Q.empty())
        {
            n1=Q.front();Q.pop();
            if(n1.x==b)
            {
                res=n1.t;
                break;
            }
            n2.x=n1.x-1;n2.t=n1.t+1;
            if(n2.x==b){res=n2.t;break;}
            else if(n2.x>0&&n2.x<1000000&&mark[n2.x])
            {
                Q.push(n2);
                mark[n2.x]=false;
            }
            n2.x=n1.x+1;n2.t=n1.t+1;
            if(n2.x==b){res=n2.t;break;}
            else if(n2.x>0&&n2.x<1000000&&mark[n2.x])
            {
                Q.push(n2);
                mark[n2.x]=false;
            }
            n2.x=n1.x*2;n2.t=n1.t+1;
            if(n2.x==b){res=n2.t;break;}
            else if(n2.x>0&&n2.x<1000000&&mark[n2.x])
            {
                Q.push(n2);
                mark[n2.x]=false;
            }
        }
        cout<<res<<endl;
    }
    return 0;
}