bfs—Catch That Cow—poj3278

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 87152		Accepted: 27344

http://poj.org/problem?id=3278

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.

 

简单的bfs，题目是单测试例，写的是多测试例。

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;

const int MAX=100000;
int pace[MAX+10];
int main()
{
    int n,k;
    while(cin>>n>>k)
    {
        memset(pace,0,sizeof(pace));
        int t;
        queue<int> qu;
        qu.push(n);
        while(!qu.empty())
        {
            t=qu.front();
            if(t==k)
            {
                cout<<pace[t]<<endl;
                return 0;
            }
            qu.pop();
            if(t<MAX&&!pace[t+1])
            {
                qu.push(t+1);
                pace[t+1]=pace[t]+1;
            }
            if(t>0&&!pace[t-1])
            {
                qu.push(t-1);
                pace[t-1]=pace[t]+1;
            }
            if(t*2<=MAX&&!pace[t*2])
            {
                 qu.push(t*2);
                 pace[t*2]=pace[t]+1;
            }
        }
    }
    return 0;
}