POJ 3278:Catch That Cow 抓住那头牛
POJ 3278:Catch That Cow 抓住那头牛
- 总时间限制:
- 2000ms
- 内存限制:
- 65536kB
- 描述
-
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?
- 输入
- Line 1: Two space-separated integers: N and K
- 输出
- Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
- 样例输入
-
5 17
- 样例输出
-
4
- 提示
- 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.
题目分析：
Finding the shortest path，采用广度搜索BFS的方法。
注意分析题目：
Farmer John 的位置是N， cow 牛的位置是 K；
action只有3种，+1，-1，*2； 没有 “除以2”。
所以当N>K，时，只能执行-1 action。 即 如N-K，最短steps= K-N。
当N
N>0; 可以执行-1 action。
N
2*N
样例代码：
/*
广度优先搜索，重要的是剪枝和标记访问过的路径。
所谓的剪枝，也可以理解为生成每一个子节点的条件；所谓条件限制越严谨，剪枝效果越好。
*/
#include
#include
#include
using namespace std;
const int MAX=100020; // 这个MAX的设定并不太严谨
int visited[MAX]={0};
int main()
{
int n,k;
while (cin>>n>>k){
if( n > k ){
cout< q;
q.push(n);
int t=0;
while( !q.empty() ){
t = q.front();
q.pop();
if(t == k){
break;
}
if( t < k && !visited[t+1] ){ // 当t>k时，不执行+1 action
q.push(t+1);
visited[t+1] = visited[t] + 1;
}
if( t > 0 && !visited[t-1] ){ //
q.push(t-1);
visited[t-1] = visited[t] + 1;
}
if( 2*t < MAX && !visited[2*t] ){ //此处不太严谨，如上讨论
q.push(2*t);
visited[2*t] = visited[t] +1;
}
}
cout<