poj 3278 BFS

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 30928		Accepted: 9537

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.

 

---

一般的广搜，不过要考虑的就是不要让数组越界

 

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

struct M{
    int add;
    int step;
};

M queue[2000005];
bool vis[2000005]={false};

int BFS(int s, int e)
{
    
    int head=0;
    int tail=0;
    queue[0].add=s;
    queue[tail++].step=0;
    while(head<tail)
    {
        M x=queue[head++];
        if(x.add==e)
        {
            return x.step;
        }
        if(!vis[x.add+1])
        {
            vis[x.add+1]=true;
            queue[tail].add=x.add+1;
            queue[tail++].step=x.step+1;
        }
        if( x.add-1>=0 && !vis[x.add-1])    //x.add-1>=0 这个不要忘了，不然会runtime error
        {
            vis[x.add-1]=true;
            queue[tail].add=x.add-1;
            queue[tail++].step=x.step+1;
        }
        if(!vis[x.add*2] && x.add*2<=200000)  // 同样的，add*2<=200000也是必须的，否则由于广搜，会乘到很大。
        {
            vis[x.add*2]=true;
            queue[tail].add=x.add*2;
            queue[tail++].step=x.step+1;
        }
    }
    return 0;
}

int main()
{
    int N,K;
    int i,j;
    while(scanf("%d%d",&N,&K)!=EOF)
    {
        printf("%d\n",BFS(N,K));
    }
    return 0;
}