（bfs）POJ3278 Catch That Cow

最新推荐文章于 2019-06-12 11:59:00 发布

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最新推荐文章于 2019-06-12 11:59:00 发布

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分类专栏：图论搜索 - BFS 文章标签：数据结构-图

本文链接：https://blog.youkuaiyun.com/qq_37275680/article/details/77318403

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本文介绍了一个有趣的算法问题——如何利用步行和瞬移两种方式，在最短时间内追上静止不动的牛。通过广度优先搜索(BFS)算法实现，探讨了不同移动策略下达到目标位置所需的最短时间。

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Catch That Cow

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.

Example Input

5 17

Example Output

解题思路
两种交通方式，三种走法，即点可能存在的三种邻接点。
后记，代码1是最初最这个题的代码，2018/5/14再来补个代码，见代码2

代码1

#include<stdio.h>
#include<cstring>
#include<queue>
using namespace std;
typedef struct Node{
    int num,step;
}Node;
int visited[100001],k;
int Check(int t)//检查结点的合法性
{
    if(t<0||t>100000||visited[t]==1)
    /*注意点：if(visited[t]==1||t<0||t>100000)这种情况就报错，
    如果t<0||t>100000，visited[t]就存在越界错误*/
        return 0;
    return 1;
}
void BFS(int n)
{
    queue<Node>q;
    Node fir;
    fir.num=n;
    fir.step=0;
    visited[n]=1;
    q.push(fir);
    while(!q.empty())
    {
        Node tmp,nxt;
        tmp=q.front();
       // printf("num=%d\n",tmp.num);
        q.pop();
        if(tmp.num==k)
        {
            printf("%d\n",tmp.step);
            return;
        }
        nxt.num=tmp.num+1;
        if(Check(nxt.num))
        {
            nxt.step=tmp.step+1;
            visited[nxt.num]=1;
            q.push(nxt);
        }

         nxt.num=tmp.num-1;
        if(Check(nxt.num))
        {
            nxt.step=tmp.step+1;
            visited[nxt.num]=1;
            q.push(nxt);
        }

         nxt.num=tmp.num*2;
        if(Check(nxt.num))
        {
            nxt.step=tmp.step+1;
            visited[nxt.num]=1;
            q.push(nxt);
        }
    }
}
int main()
{
    int n;
    scanf("%d%d",&n,&k);
    memset(visited,0,sizeof(visited));
    BFS(n);
    return 0;
}

代码2

#include<iostream>
#include<cstdio>
#include<string>
#include<string.h>
#include<queue>
using namespace std;
const int maxn=1e5+1;
typedef struct Node{
    int x,step;
}Node;
int vis[maxn];

int bfs(int s,int t)
{
    if(s==t)
        return 0;
    memset(vis,0,sizeof(vis));
    queue<Node> q;
    Node tmp,nxt;
    tmp.x=s;
    tmp.step=0;
    q.push(tmp);
    vis[s]=1;
    while(!q.empty()){
        tmp=q.front();
        nxt.step=tmp.step+1;
        q.pop();
        for(int i=0;i<3;i++)
        {
            if(i==0)
                nxt.x=tmp.x-1;
            else if(i==1)
                nxt.x=tmp.x+1;
            else
                nxt.x=tmp.x*2;
            if(nxt.x<0||nxt.x>=maxn)
                continue;
            if(!vis[nxt.x]){
                if(nxt.x==t)
                    return nxt.step;
                vis[nxt.x]=1;
                q.push(nxt);
            }
        }
    }
    return 0;
}

int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    printf("%d\n",bfs(n,k));
    return 0;
}