POJ训练计划3278_Catch That Cow/SDUT2782_我跳我跳我跳跳跳（一维BFS）

最新推荐文章于 2020-04-15 13:49:11 发布

原创最新推荐文章于 2020-04-15 13:49:11 发布 · 686 阅读

0 ·

CC 4.0 BY-SA版权

POJ 同时被 3 个专栏收录

99 篇文章

订阅专栏

POJ训练计划

41 篇文章

订阅专栏

SDUT

40 篇文章

订阅专栏

描述了一种在有限步数内捕获逃逸牛的方法，通过行走和瞬移两种方式在数轴上寻找目标。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 43401		Accepted: 13512

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.

我跳我跳我跳跳跳

Time Limit: 1000MS Memory limit: 65536K

题目描述

有一条直线，上有n个点，编号从0到n-1。当小A站在s点处，每次可以往前跳到s+1，也可以往前跳到s-1(当s-1 >= 0时)，也可以调到2*s处。现在问小A最少跳多少次才能跳到点e处。

输入

多组输入。每组输入两个整数s，e（0 <= s,e <= 100,000）。n趋于无穷大。

输出

输出小A从s跳到e的最小次数。

示例输入

5 17

示例输出

提示

来源

解题报告

比赛的时候根本没有想到是bfs解题的，还以为是背包呢，bfs一维。。。

多组输入每次都是忘了数组清零的节奏。。。

#include <iostream>
#include <stdio.h>
#include <string.h>


using namespace std;
struct node
{
    int x,num;
}q[1000000];
int s,e;
int v[1000000];
void bfs()
{
    node now,next;
    int t=0,f=0;
    now.x=s;
    now.num=0;
    q[f++]=now;
    v[now.x]=1;
    while(t<f)
    {

        now=q[t++];
        //printf("x =%d,num =%d\n",now.x,now.num);
        if(now.x==e)
        {
            printf("%d\n",now.num);
            return ;
        }
        next.x=now.x+1;
        if(next.x>=0&&next.x<300000&&v[next.x]!=1 )
        {
            next.num=now.num+1;
            q[f++]=next;
            v[next.x]=1;
        }
        next.x=now.x-1;
        if(next.x>=0&&next.x<300000&&v[next.x]!=1)
        {
            next.num=now.num+1;
            q[f++]=next;
            v[next.x]=1;
        }
        next.x=now.x*2;
        if(next.x>=0&&next.x<300000&&v[next.x]!=1)
        {
            next.num=now.num+1;
            q[f++]=next;
            v[next.x]=1;
        }
    }
}
int main(){
    while(~scanf("%d%d",&s,&e)){
        memset(v,0,sizeof(v));
        bfs();
    }
}