POJ3278 Catch That Cow(广搜BFS)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 54969		Accepted: 17198

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.

题目大意:

给定一个初始值m,一个终值n，有三种方法分别为m + 1,m - 1,m * 2，求用最小的步数刚好等于n。

解题思路:

最小距离，想到了广搜，广搜用队列解就可以，由于元素可能为0，所以m-1可能出现-1的情况，然而数组下标不能为-1,然后就开始一直RE,在POJ discuss上发现有人也是被坑了，果断改了，然后就AC了。

AC代码:

#include<iostream>
#include<queue>
using namespace std;

const int maxn = 200010;
int step[maxn];
bool vis[maxn];
int m,n;

void bfs()
{
int start = m;
int final = n;
queue<int> que;
que.push(start);
vis[start] = true;
while(!que.empty())
{
int element = que.front();
que.pop();
if(element + 1< maxn && !vis[element + 1]) //如果不越界又没有标记过，就加入队列
{
que.push(element + 1);
vis[element + 1] = true;
step[element + 1] = step[element] + 1; //当前这个状态(element)到(element+1)一共经历了一步
}
if(element - 1< maxn && !vis[element - 1] && element != 0)  //此处要特别注意,如果element刚好为0的时候，数组下标为-1,re.
{
que.push(element - 1);
vis[element - 1] = true;
step[element - 1] = step[element] + 1; //当前这个状态(element)到(element-1)一共经历了一步
}
if(element * 2 < maxn && !vis[element * 2])
{
que.push(element * 2);
vis[element * 2] = true;
step[element * 2] = step[element] + 1; //当前这个状态(element)到(element*2)一共经历了一步
}
if(vis[final])
{
cout<<step[final]<<endl;
break;
}
}
}

int main()
{
int i;
while(cin>>m>>n)
{
if(m == n)
{
cout<<"0"<<endl;
return 0;
}
else
{
for(i=0;i<=200000;i++)
{
step[i] = 0;
vis[i] = false;
}
}
bfs();
}
return 0;
}