[kuangbin带你飞]专题一 简单搜索-C - Catch That Cow POJ - 3278
Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 108204 | Accepted: 33803 |
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.
题解：
题目的意思是牛在一个坐标K的地方，农夫在坐标N的地方，农夫有三种方式去行走，（1）N*2 （2）N+1 （3）N-1
想要最短的方式走到，显然是用宽搜的方式，但是重点来了，宽搜也是有技巧的。
首先一定要定义一个vis来储存有没有走过这个点。这是最最必要的优化，否则会有死循环的。
另外，有关于题目方面的优化也很重要，这道题，我可以总结出三个优化方案。没有优化剪枝，程序会非常的繁琐。
再者，使用 qu.empty() 确实要比 qu.size() 要快上很多。
#include
#include
using namespace std;
int N,K;
struct Node {
int Lo,step;
Node(int Lo = 0,int step = 0):Lo(Lo),step(step){};
};
queue qu;
int vis[200007];
int bfs(){
qu.push(Node(N,0));
Node t;
while (!qu.empty()){
t = qu.front();
qu.pop();
int Lo = t.Lo;
int step = t.step;
if (Lo == K) break;
if (Lo > K*2 || Lo > 200000 || Lo < 0)continue;
if (vis[Lo])continue;
vis[Lo] = 1;
qu.push(Node(Lo*2,step+1));
qu.push(Node(Lo+1,step+1));
qu.push(Node(Lo-1,step+1));
}
return t.step;
}
int main()
{
cin >> N >> K;
if (N >= K) cout << N-K << endl;
else {
int ans = bfs();
cout << ans << endl;
}
}