poj3278: Catch That Cow结题报告

Catch That Cow Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 96169 Accepted: 30168

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.

Source

USACO 2007 Open Silver

代码：

#include<cstdio> #include<queue> #include<cstring> using namespace std; bool visit[200002]; typedef pair<int, int> PAIR; int bfs(int n, int k){ queue<PAIR> que; que.push(PAIR(n, 0)); if(n == k) return 0; visit[n] = true; while(!que.empty()){ PAIR p = que.front(); que.pop(); if(p.first == k) return p.second; if(p.first - 1 >= 0 &&!visit[p.first - 1]){ que.push(PAIR(p.first - 1, p.second + 1)); visit[p.first - 1] = true; } if(p.first + 1 <= 100000 && !visit[p.first + 1]){ que.push(PAIR(p.first + 1, p.second + 1)); visit[p.first + 1] = true; } if((p.first<<1) <= 100000 && !visit[p.first<<1]){ que.push(PAIR(p.first<<1, p.second + 1)); visit[p.first<<1] = true; } } return -1; } int main(){ int n, k; scanf("%d%d",&n,&k); memset(visit, 0, sizeof(visit)); int ans = bfs(n, k); printf("%d\n",ans); return 0; }