总时间限制: 2000ms 内存限制: 65536kB

描述

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?

输入

Line 1: Two space-separated integers: N and K

输出

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

样例输入

5 17

样例输出

4

思路

这是一个求最少步数的问题, 特别适合用广度优先搜索解决.

改进后完整代码如下：

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
int N, K;
const int MAXN = 100000;    // 要开很大的数组才行
int visited[MAXN + 10];     // 判重标记, visited[i] = true表示i已扩展
typedef struct Step {
    int x;          // 位置
    int steps;      // 到达x所需的步数
    Step(int xx, int s): x(xx), steps(s) {}
}Step;
queue<Step> OpenQueue;  // Open队列
int main() {
    /* Read data and initialize them */
    cin >> N >> K;
    memset(visited, 0x0, sizeof(visited));
    OpenQueue.push(Step(N, 0));
    visited[N] = 1;
    /* BFS */
    while( !OpenQueue.empty() ) {
        Step s = OpenQueue.front();
        if( s.x == K ) {    // 找到目标
            cout << s.steps << endl;
            return 0;
        } else {
            if( s.x - 1 >= 0 && !visited[s.x-1] ) {     // 左走一步
                OpenQueue.push(Step(s.x-1, s.steps+1));
                visited[s.x-1] = 1;
            }
            if( s.x + 1 <= MAXN && !visited[s.x+1] ) {  // 右走一步
                OpenQueue.push(Step(s.x+1, s.steps+1));
                visited[s.x+1] = 1;
            }
            if( s.x*2 <= MAXN && !visited[2 * s.x] ) {    // 横跳
                OpenQueue.push(Step(2 * s.x, s.steps+1));
                visited[2 * s.x] = 1;
            }
            OpenQueue.pop();
        }
    }
    return 0;
}