POJ3278 Catch That Cow(BFS)

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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.

一个广搜题目，首先我们应该判断条件是否否和，则需要找出判断条件（边界条件）：

1.x-1>=0，因为一旦x-1小于0了，则乘二倍肯定是小于0的，不满足了；

2.x+1<=k，一旦超过了k，那则是做了无用功；

3.x*2<=100000，一个很好知道的判断条件；

满足了这些则是简单明了的广搜了，直接搞就行。

/* ***********************************************
Author        :Mosu
Created Time  :2015/10/10 18:30:45
File Name     :\Users\Mosu\Desktop\project\BC.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7ffffff
struct Node{
int x;
int step;
};

bool vis[1000009];

int main()
{
int n,k;
while(scanf("%d%d",&n,&k)==2){
queue<Node>q;
memset(vis,0,sizeof(vis));
while(!q.empty())
q.pop();
Node temp;
temp.x=n;
temp.step=0;
q.push(temp);
while(!q.empty()){
Node now=q.front();
q.pop();
if(now.x==k){
printf("%d\n",now.step);
break;
}
if(now.x-1>=0&&!vis[now.x-1]){
vis[now.x-1]=true;
temp.x=now.x-1;
temp.step=now.step+1;
q.push(temp);
}
if(now.x+1<=k&&!vis[now.x+1]){
vis[now.x+1]=true;
temp.x=now.x+1;
temp.step=now.step+1;
q.push(temp);
}
if(now.x<=k&&now.x*2<=100000&&!vis[now.x*2]){
vis[now.x*2]=true;
temp.x=now.x*2;
temp.step=now.step+1;
q.push(temp);
}
}
}
return 0;
}