poj 3278 Catch That Cow 广搜

hdu 2717 Catch That Cow,题目链接

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 11466 Accepted Submission(s): 3551

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.

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.

Source

USACO 2007 Open Silver

Recommend

teddy | We have carefully selected several similar problems for you: 2102 1372 1240 1072 1180

题意：给你两个位置k和n，求出从k到n的最小步数。

从k有三种方式变化方式： k+1，k-1，k*2。

队列广搜，一个vis数组查看是否访问过，一个step数组记录步数。

#include<stdio.h>
#include<queue>
#include<algorithm>
#include<string>
#include<iostream>
#include<string.h>
#include<math.h>
#define maxn 100005
using namespace std;
int vis[maxn];//标记是否访问过;
int step[maxn];//记录步数;
queue <int > q ;
int serch(int k,int n)
{
    int flag=0,a,mark=0;
    memset(vis,0,sizeof(vis));//数组清零;
    memset(step,0,sizeof(step));//数组清零;
    q.push(k);  //将k入队;
    vis[k]=1;   //标记为1;
    while(!q.empty())   //当数组非空时;
    {
        a=q.front();    //将队列首位赋值给a;
        q.pop();        //删除队列首位;
        for(int i=1; i<=3; i++)
        {
            if(i==1)
                flag=a+1;
            else if(i==2)
                flag=a-1;
            else
                flag=a*2;
            if(flag>100000||flag<0)//越界判断;
                continue;
            if(!vis[flag])//如果标记为0;
            {
                step[flag]=step[a]+1;//当前步数等于上次步数+1;
                vis[flag]=1;//标记为1;
                q.push(flag); //入队;
                if(flag==n)//如果当前值为n;
                    return step[flag];//返回步数;
            }
        }
    }
    return 0;
    //如果是提交C++的话，要不要返回值都可以;
    //但如果是G++的话，必须要一个return 0;
    //这个是编译器的问题，具体也不是很清楚;
}
int main()
{
    int n,k;        //k追n;
    while(~scanf("%d%d",&k,&n))//在这里我写成了k追n;
    {
        if(k>n) //如果k>n，只能通过不断减一得到，也就是k-n步;
            printf("%d\n",k-n);
        else
            printf("%d\n",serch(k,n));
    }
    return 0;
}

之前关于那个return 0，wa了，好坑，换了编译器就A了。

T^T，刚刚问了远航学长关于那个C++提交，G++提交不同的问题，结果学长也不知道，我也是第一次遇到这种问题，好忧桑~~

不过学长也说没见过，他说只是在这个oj上遇到的是 double要用 %.f 好坑的OJ~~QAQ