隐式图--HDU - 2717 Catch That Cow

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. 

分析：隐式图，把每个点的x坐标看成是一种状态，依题有3种状态转移方式；每转移一步，vis+1；求最短路径用BFS

#include <cstdio>#include <cstring>#include <queue>using namespace std;const int maxn = 100000 + 10;int vis[maxn], n, k;void bfs(){queue<int> q;q.push(n);while(!q.empty()) {int u = q.front(); q.pop();        if(u == k) {printf("%d\n", vis[k]);return ;        }        if(!vis[u+1]) {vis[u+1] = vis[u] + 1;q.push(u+1);        }        if(!vis[u-1]) {vis[u-1] = vis[u] + 1;q.push(u-1);        }        if(2*u<maxn && !vis[2*u]) {vis[2*u] = vis[u] + 1;q.push(2*u);        }}}int main(){while(~scanf("%d%d", &n, &k)) {memset(vis, 0, sizeof(vis));bfs();}    return 0;}