poj3278 Catch That Cow

本来用DFS做的，总是超时，拿别人AC的程序一测数据结果还一样，但是就是超时，这道题用DFS做要有很多剪枝，不然会有很多的重复，比如第x步走过点m，回溯后再DFS第y步的时候又走过m点，会有大量重复情况，还是BFS比较好剪枝

/*
Catch That Cow
Time Limit: 2000MS      Memory Limit: 65536K
Total Submissions: 63279        Accepted: 19813
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 <cstdio>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <limits.h>
#include <algorithm>
#include <queue>

using namespace std;

const int maxn = 100010;
bool vis[maxn];
int sizev = sizeof(vis);
int step[maxn];    ///用于记录走过一个点时是第几步
queue<int> q;

int bfs(int N, int K) {
    while (!q.empty()) q.pop();   ///使用前清空队列

    q.push(N);    ///将FJ所在的点入队
    vis[N] = true;
    step[N] = 0;

    int head, next;

    while (!q.empty()) {    ///队列不为空时BFS
        head = q.front();
        q.pop();
        for (int i = 0; i < 3; i++) {    ///对于三个方向BFS
            if (i == 0) {
                next = head + 1;
            } else if (i == 1) {
                next = head - 1;
            } else if (i == 2) {
                next = head << 1;
            }
            if (next < 0 || next > 100000 || vis[next]) continue;   ///若越界或以前标记过则continue
                                ///其中对vis[next]判断剪枝，用DFS的话似乎不太好标记之前是否重复走过

            q.push(next);     ///将新的点入队
            vis[next] = true;
            step[next] = step[head] + 1;    ///走到next用了step[head] + 1步

            if (next == K) return step[next];   ///当走到K时此时step[next]既是最少步数，直接返回
        }
    }
}

int main()
{
    int N, K;
    while (~scanf("%d%d", &N, &K)) {
        if (N >= K) {              ///若N>=K则只能每次N - 1这样走，所以直接减就行了
            printf("%d\n", N - K);
            continue;
        }
        memset(vis, false, sizev);
        printf("%d\n", bfs(N, K));
    }
    return 0;
}