POJ 3278-- Catch That Cow

题目：

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total 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 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));
}
}