題目網址：http://poj.org/problem?id=3278

題目：

Catch That Cow

Time Limit:?2000MS	?	Memory Limit:?65536K
Total Submissions:?92360	?	Accepted:?28999

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.

思路：?

很簡單的一道BFS題。總共只有三種操作，每次都取隊首元素分別進行三種操作，再對操作結果進行判斷，篩去負數和超過100,000（邊界值）的情況，其他情況都入隊。

代碼：

1 #include <cstdio> 2 #include <queue> 3 #include <algorithm> 4 using namespace std; 5 int visited[100005]; 6 int n,k; 7 queue<int>q; 8 void bfs(){ 9 q.push(n); 10 while (!q.empty()) { 11 int x=q.front();q.pop(); 12 if(x==k){ 13 printf("%d\n",visited[x]-1);//初始次數為1，結果減去1 14 return ; 15 } 16 if(x-1>=0 && !visited[x-1]){//負數不考慮 17 q.push(x-1); 18 visited[x-1]=visited[x]+1; 19 } 20 if(x+1<=100000 && !visited[x+1]){//超過邊界值的要篩去，否則會占用不必要的搜索時間 21 q.push(x+1); 22 visited[x+1]=visited[x]+1; 23 } 24 if(x*2<=100000 && !visited[x*2]){ 25 q.push(x*2); 26 visited[x*2]=visited[x]+1; 27 } 28 } 29 } 30 int main(){ 31 scanf("%d%d",&n,&k); 32 visited[n]=1;//初始位置也要賦值，防止第二次搜索到 33 bfs(); 34 return 0; 35 }

?

轉載于:https://www.cnblogs.com/uniles/p/7145860.html

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