HDU2717 Catch That Cow

Problem 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 minuteTeleporting: 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.

思路

给定两个数，n，k，求从n到k所走的次数所少，每次可以加1，或减1或翻倍。

简单bfs。

代码

#include<algorithm> #include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #include<iostream> #include<queue> using namespace std; long int n, k; int dir[2] = {-1, 1}; long int t; long int now; int visit[100010]; int step[100010]; int bfs() { queue<long int> q; q.push(n); visit[n] = 1; while(!q.empty()) { t = q.front(); q.pop(); if(t == k) return 0; for(int i = 0; i < 3; i++) { if(i == 2) { now = t * 2; } else now = t + dir[i]; if(now < 0 || now > 100000) continue; if(!visit[now]) { q.push(now); visit[now] = 1; step[now] = step[t] + 1; } if(now == k) return step[now]; } } return -1; } int main() { while(scanf("%ld%ld",&n,&k) != EOF) { memset(visit, 0, sizeof(visit)); memset(step, 0, sizeof(step)); printf("%d\n",bfs()); } return 0; }