poj3278 Catch That Cow

Catch That Cow
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 89585 Accepted: 28099
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n到k，有三种移动方法，-1，+1，*2；
dfs三种走法，
注意：1.减枝
2.vis标记是否走过，step记录步数

#include<stdio.h>
#include<queue>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int n,k;
int ans;
int vis[100001],step[100001];
queue<int>c;
int dfs(int n,int ans)
{
    vis[n]=1;
    step[n]=0;
    c.push(n);
    //cout<<"123"<<endl;
    while(!c.empty())
    {
        //cout<<"123"<<endl;
        int fist=c.front();
        c.pop();
        //printf("%d
",fist);
        if(fist==k)
            return step[fist];
        for(int i=0;i<3;i++)
        {
            int next=0;
            if(i==0) next=fist-1;
            else if(i==1) next=fist+1;
            else if(i==2) next=fist*2;
            if(next<0||next>=100001)
                continue;
            else
            {
                if(vis[next]==0)
                {
                    vis[next]=1;
                    step[next]=step[fist]+1;
                    c.push(next);
                }
            }
        }
    }
}
int main()
{
    memset(vis,0,sizeof(vis));
    memset(step,0,sizeof(step));
    while(!c.empty()) c.pop();
    scanf("%d%d",&n,&k);
    int aa=dfs(n,0);
    printf("%d
",aa);
    return 0;
}