POJ 3278 Catch That Cow(BFS)

POJ 3278  Catch That Cow

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.

 

题目大意：一条笔直路上有农夫和牛，牛的位置是固定的，农夫可以前进一步或者后退一步，或者跳到当前位置的两倍处，求最少移动次数

 

分析：bfs求最短路

 

代码：

#include<iostream>
#include<cstdio> 
#include<queue>
using namespace std;
const int M = 100000;
queue<int> que;
int N, K;
int d[M + 5] = {0};
void bfs() {
    que.push(N);
    while(que.size()) {
        int cur = que.front();
        que.pop();
        if(cur == K) {
            printf("%d\n", d[cur]);    
            break;
        }
        if(cur - 1 >= 0 && cur - 1 <= M && !d[cur-1]) {
            que.push(cur - 1);
            d[cur-1] = d[cur] + 1;
        }
        if(cur + 1 >= 0 && cur + 1 <= M && !d[cur+1]) {
            que.push(cur + 1);
            d[cur+1] = d[cur] + 1;
        }
        if(cur * 2 >= 0 && cur * 2 <= M && !d[cur*2]) {
            que.push(cur*2);
            d[cur*2] = d[cur] + 1;
        }
    }
}
int main() {
    cin >> N >> K;
    bfs();
    return 0;
}