bfs(最短路径)

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

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 146388		Accepted: 44997

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.

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdio.h>
#include <queue>
#include <stack>;
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define mod 1000000007
using namespace std;
typedef long long ll ;
int vis[1300009];
int ans[1300009];
int n , m ;

void bfs(int b)
{
    queue<int>q;
    memset(vis , 0 , sizeof(vis));
    memset(ans , 0 , sizeof(ans));
    if(n < m)
    {

        q.push(b);
        vis[b] = 1 ;
        ans[b] = 0 ;
        while(!q.empty())
        {

            int x = q.front();
            if(x == m)
            {
                printf("%d
" , ans[m]);
                break ;
            }
            q.pop() ;
            if(x >= 0 && x <= 100000 && !vis[x*2])
            {
                vis[2*x] = 1 ;
                ans[2*x] += vis[x] + ans[x];
                q.push(2*x);
            }
            if(x >= 0 && !vis[x+1])
            {
                vis[x+1] = 1 ;
                ans[x+1] += vis[x] + ans[x];
                q.push(x+1);
            }
            if(x - 1 >= 0 && !vis[x-1])
            {
                vis[x-1] = 1 ;
                ans[x-1] += vis[x] + ans[x];
                q.push(x-1);
            }
        }

    }
    else
    {
        printf("%d
" , n - m);
    }

}


int main()
{

    while(~scanf("%d%d" , &n , &m))
    {
        bfs(n);
    }

    return 0;
}