POJ 3278-- Catch That Cow

题目：

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 76114 Accepted: 24036

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX + 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 andK

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.

题意：
FJ要抓牛，牛在k位置，FJ在n位置，假设牛是在原地不动的，FJ假设在X点，可以向X+1走一步，或者X-1走一步，或者2*X走一步，问最短路。

思路：
如果牛在FJ后面，那么FJ只能一步一步向后走，否则宽搜。

实现：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <math.h>#include <queue>using namespace std;const int maxn = 100005;struct node {    int nn, step;};int _map[maxn];//记录到某一点的最短步数int bfs (int n, int k) {    queue <node>q;    node head;    head.nn = n;    head.step = 0;    _map[head.nn] = 0;    q.push(head);    while(!q.empty()) {        node en;        node start = q.front();        if (start.nn == k)            return _map[k];        q.pop();        for (int i = 0; i < 3; i++) {            if (i == 0 && start.nn + 1 <= maxn) {                en.nn = start.nn + 1;                en.step = start.step + 1;            }            else if (i == 1 && start.nn - 1 >= 0) {                en.nn = start.nn - 1;                en.step = start.step + 1;            }            else if (i == 2 && 2 * start.nn <= maxn) {                en.nn = start.nn * 2;                en.step = start.step + 1;            }            if (_map[en.nn] > start.step + 1) {                _map[en.nn] = start.step + 1;                q.push(en);            }        }    }return -1;}int main() {    int n, k;    while (scanf("%d%d", &n, &k) != EOF) {        for (int i = 0; i <= maxn; i++) {            _map[i] = maxn + 5;        }        if (n >= k)            printf("%d\n", n - k);        else            printf("%d\n", bfs(n, k));    }}