POJ 3278Catch That Cow （bfs）

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 95484 Accepted: 29934

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.

题意：农夫在位置n，他可以走x-1,x+1,2*x.最小走多少步可以找到位置在k的牛

做法：bfs

要注意的地方有两个，边界条件，和结构体的入队。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>using namespace std;#define MAX_N 100010struct john{int x;int t;};int main(){int n,k;while (cin >> n >> k) {if (n >= k)     //比他小的情况和等于的情况直接就可以求出来了{cout << n-k<< endl;continue;}//初始化queue<john>q;john start;int v[MAX_N] = {0};start.x = n;start.t = 0;v[n] = 1;q.push(start);while (!q.empty()){if (q.front().x == k)break;//找到就退出去if (q.front().x + 1 < MAX_N) {   //边界一定要考虑 不然会runtime errorif (v[q.front().x + 1] == 0){john temp;//要重新定义一个变量  因为queue的对象是引用，如果不重新定义，会修改了队列里的数据v[q.front().x + 1] = 1;temp.x = q.front().x + 1;temp.t = q.front().t + 1;q.push(temp);}}if (q.front().x - 1 >= 0) {if (v[q.front().x - 1] == 0){john temp;v[q.front().x - 1] = 1;temp.x = q.front().x - 1;temp.t = q.front().t + 1;q.push(temp);}}if (q.front().x * 2 < MAX_N){if(v[q.front().x * 2] == 0){john temp;v[q.front().x * 2] = 1;temp.x = q.front().x * 2;temp.t = q.front().t + 1;q.push(temp);}}q.pop();}cout << q.front().t << endl;}return 0;}