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?

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

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

5 17

4

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);elseprintf("%d\n", bfs(n, k));}
}