POJ3278 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.

这明显是一道最短路径的题，用bfs即可，从三个方向进行广搜，队列大小设置足够大，便可以了。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
#include<vector>
#include<stack>
#include<queue>
using namespace std;

int n, k, t[100010];

int bfs(){
    memset(t, -1, sizeof(t));
    queue<int> que;
    que.push(n);
    t[n] = 0;
    int temp;
    while(!que.empty()){
        temp = que.front();
        que.pop();
        if(temp==k)
            break;
        int next;
        next = temp - 1;
        if(next>=0&&next<=100000&&t[next]==-1){
            que.push(next);
            t[next] = t[temp] + 1;
        }
        next = temp + 1;
        if(next>=0&&next<=100000&&t[next]==-1){
            que.push(next);
            t[next] = t[temp] + 1;
        }
        next = temp * 2;
        if(next>=0&&next<=100000&&t[next]==-1){
            que.push(next);
            t[next] = t[temp] + 1;
        }
    }
    return t[temp];
}

int main(){
    while(~scanf("%d %d",&n,&k)){
        cout << bfs();
    }
    return 0;
}