hdoj 2717 Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8095    Accepted Submission(s): 2565

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.
 

广搜ac，要注意开的数组范围以及中间的一些判定。

 

 

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define max 1000000+10
using namespace std;
int visit[max],step[max];
void bfs(int start,int end)
{
    int now,next;
    int i,j;
    queue<int>q;
    q.push(start);
    visit[start]=1;
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        if(now==end) break;
        next=now+1;
        if(next<=end&&!visit[next])
        {
            step[next]=step[now]+1;
            visit[next]=1;
            q.push(next);
        }
        next=now-1;
        if(next>=0&&!visit[next])
        {
            step[next]=step[now]+1;
            visit[next]=1;
            q.push(next);
        }
        next=now*2;
        if(next<=max&&!visit[next])
        {
            step[next]=step[now]+1;
            visit[next]=1;
            q.push(next);
        }
    }
    printf("%d\n",step[end]);
}
int main()
{
    int i,j;
    int start,end;
    while(scanf("%d%d",&start,&end)!=EOF)
    {
        memset(visit,0,sizeof(visit));
        memset(step,0,sizeof(step));
        bfs(start,end);
    }
    return 0;
}