Problem Descrption

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.

問題鏈接：http://poj.org/problem?id=3278

問題分析：搜索類問題，首先想到DFS和BFS兩種，像這種很大的，如果用深搜很容易超時(shí)，所以明顯是BFS，一開始寫

為了記錄步數(shù)使用了結(jié)構(gòu)數(shù)組，然而這樣會ML，這也給我提醒，以結(jié)構(gòu)類型的隊(duì)列可能內(nèi)存較大，后來利用及數(shù)組記錄父子節(jié)點(diǎn)的方式解決，可以建立數(shù)組ed[tx]=t，即記錄tx的父節(jié)點(diǎn)為t，最后再回溯即可；

AC代碼：

#include<iostream> #include<queue> #include<cstring> using namespace std; int n, k; int vis[100005]; int ed[100005]; void bfs() {queue<int>q;q.push(n);vis[n];while (!q.empty()){int t = q.front();if (t == k){return;}q.pop();int tx;for (int i = 0; i < 3; i++){if (i == 0) { tx = t + 1; }if (i == 1) { tx = t * 2; }if (i == 2) { tx = t - 1; }if (tx > 100000 || tx < 0 || vis[tx])continue;q.push(tx);ed[tx] = t;vis[tx]=1;}} } int main() {while (cin >> n >> k){memset(vis, 0, sizeof(vis));bfs();int t=k,step=0;while (t!=n){t = ed[t];step++;}cout << step << endl;} }

因使用結(jié)構(gòu)隊(duì)列而ML的代碼：

#include<iostream> #include<queue> #include<cstring> using namespace std; int n, k,min1=999999; int vis[100005]; struct way {int x;int step; }; void bfs() {queue<way>q;struct way a;a.x = n, a.step = 0;q.push(a);vis[a.x];while (!q.empty()){struct way t = q.front();if (t.x == k){if (t.step < min1)min1 = t.step;return;}q.pop();struct way tx;for (int i = 0; i < 3; i++){if (i == 0) { tx.x = t.x + 1; tx.step =t.step+1; }if (i == 1) { tx.x = t.x * 2; tx.step =t.step+1; }if (i == 2) { tx.x = t.x - 1; tx.step = t.step + 1; }if (tx.x > 100000 || tx.x < 0 || vis[tx.x])continue;q.push(tx);vis[tx.x];}} } int main() {while (cin >> n >> k){memset(vis, 0, sizeof(vis));min1 = 9999999;bfs();cout << min1 << endl;} }

?

總結(jié)

以上是生活随笔為你收集整理的POJ3278（BFS入门）的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。