POJ3278 Catch That Cow BFS入门

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

为求解最短路径，用BFS效果更好。

#include <iostream>#include <cstdio>#include <map>#include <set>#include <vector>#include <queue>#include <stack>#include <cmath>#include <algorithm>#include <cstring>#include <string>using namespace std;#define INF 0x3f3f3f3fconst int maxn=100000;typedef long long LL;int n,k;bool v[maxn+5];struct step{    int x;    int steps;    step(int xx,int s):x(xx),steps(s){}};queue<step> q;int main(){    while(~scanf("%d%d",&n,&k)){        memset(v,false,sizeof(v));        q.push(step(n,0));        v[n]=true;        while(!q.empty()){            step s=q.front();            if(s.x==k){                printf("%d\n",s.steps);                return 0;            }            if(s.x-1>=0&&!v[s.x-1]){                q.push(step(s.x-1,s.steps+1));                v[s.x-1]=true;            }            if(s.x+1<=maxn&&!v[s.x+1]){                q.push(step(s.x+1,s.steps+1));                v[s.x+1]=true;            }            if(2*s.x<=maxn&&!v[2*s.x]){                q.push(step(2*s.x,s.steps+1));                v[2*s.x]=true;            }            q.pop();        }    }    return 0;}