HDU2717 Catch That Cow BFS

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=2717

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15090    Accepted Submission(s): 4545

Problem 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.
 

 

#include <iostream>#include <cstring>#include <queue>using namespace std;const int MAXN = 1e5 + 7;int p[MAXN];void bfs(int start, int end){    queue<int> q;    q.push(start);    p[start] = 0;    while(!q.empty()){        int top = q.front();        q.pop();        if(top == end)            break;        if(0 <= top - 1 && !p[top - 1]){            q.push(top - 1);            p[top - 1] = p[top] + 1;        }        if(top + 1 < MAXN && !p[top + 1]){            q.push(top + 1);            p[top + 1] = p[top] + 1;        }        if(top * 2 < MAXN && !p[top * 2]){            q.push(top * 2);            p[top * 2] = p[top] + 1;        }    }}int main(){    int n, k;    while(cin >> n >> k){        memset(p, 0, sizeof(p));        bfs(n, k);        cout << p[k] << endl;    }}