POJ 3278 Catch That Cow（BFS）

Catch That Cow

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

题目大意：FJ的牛逃跑了，他想立即抓住它。FJ和牛都在一个数轴上，FJ在位置N，牛在位置K。FJ每分钟可以移动到N - 1、N + 1 或 2 * N 处，而牛却不会动。给出FJ和牛的位置，求FJ抓住牛所需要的最短时间。
解题思路：将每次FJ可能的移动情况入队，BFS即可。注意剪枝，先判断是否超出范围再判断是否访问过，如果顺序反过来的话，数组要开200000+，否则会RE。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int maxn = 100005;

int posn,posk;
int t[maxn];

int bfs()
{
memset(t,-1,sizeof(t));
queue<int> que;
que.push(posn);
t[posn] = 0;

int temp;
while(que.size()){
temp = que.front();
que.pop();
if(temp == posk){
break;
}

int next;
next = temp - 1;
if(0 <= next && next <= maxn - 5 && t[next] == -1){
que.push(next);
t[next] = t[temp] + 1;
}

next = temp + 1;
if(0 <= next && next <= maxn - 5 && t[next] == -1){
que.push(next);
t[next] = t[temp] + 1;
}

next = temp + temp;
if(0 <= next && next <= maxn - 5 && t[next] == -1){
que.push(next);
t[next] = t[temp] + 1;
}
}
return t[temp];
}

int main()
{
while(scanf("%d %d",&posn,&posk) != EOF){
printf("%d\n",bfs());
}
return 0;
}