POJ3278 Catch That Cow —— BFS

题目链接：http://poj.org/problem?id=3278

Catch That Cow Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 97563 Accepted: 30638

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.

Source

USACO 2007 Open Silver

题解：

一开始以为是数学找规律，但是看到数据范围很小，n<=1e5，可以用数组存；而且题目要求的是“最少步数”，而“最少步数”经常都是用BFS求的。再将BFS的思想带入看看，发现可以解决问题。

在第一次提交时，RUNTIME ERROR了。但是再看看代码，数组没有开小，不是数组的问题。后来发现再判断某个位置是否vis时，先判断了是否vis，然后再判断这个位置是否合法，这样会导致数组溢出，所以问题就出现在这里了，需谨慎！！

代码如下：

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <vector> #include <queue> #include <stack> #include <map> #include <string> #include <set> #define ms(a,b) memset((a),(b),sizeof((a))) using namespace std; typedef long long LL; const int INF = 2e9; const LL LNF = 9e18; const int MOD = 1e9+7; const int MAXN = 1e5+10; int vis[MAXN]; struct node { int val, step; }; queue<node>que; int bfs(int n, int k) { ms(vis,0); while(!que.empty()) que.pop(); node now, tmp; now.val = n; now.step = 0; vis[n] = 1; que.push(now); while(!que.empty()) { now = que.front(); que.pop(); if(now.val==k) return now.step; tmp.step = now.step+1; if(now.val+1>=0 && now.val+1<=1e5 && !vis[now.val+1] ) //先判断范围再判断vis ！！！ vis[now.val+1] = 1, tmp.val = now.val+1, que.push(tmp); if(now.val-1>=0 && now.val-1<=1e5 && !vis[now.val-1] ) vis[now.val-1] = 1, tmp.val = now.val-1, que.push(tmp); if(now.val*2>=0 && now.val*2<=1e5 && !vis[now.val*2] ) vis[now.val*2] = 1, tmp.val = now.val*2, que.push(tmp); } } int main() { int n, k; scanf("%d%d",&n, &k); cout<< bfs(n,k) <<endl; }