hdu2717(裸bfs广搜)

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题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=2717
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 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.

解题思路：

          一道很裸的bfs题。题意是给出farmer的点和cow的点，farmer可以选择每次+1、-1、*2这三种操作，问最少的操作次数可以使得farmer到达cow点。

          重新温习广搜。广搜对于求最优解或最短路很有帮助，虽然它的时间复杂度颇高，要把整棵树的所有路径都存下来，不过据说双向广搜（dbfs）可以把时间复杂度优化到O（n），加上哈希表处理的话，可以达到O（1）。

           本题考虑两种情况：1、 n < k时，有+1、*2两种走法；2、 n > k 时，只有-1。其实最后所求的最少次数相当于询问树节点为k的最短深度。

附图：

完整代码：

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <climits>
#include <cassert>
#include <complex>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")typedef long long LL;
typedef double DB;
typedef unsigned uint;
typedef unsigned long long uLL;/** Constant List .. **/ //{const int MOD = int(1e9)+7;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f3f3f3f3f3f3fLL;
const DB EPS = 1e-9;
const DB OO = 1e20;
const DB PI = acos(-1.0); //M_PI;int n , k;
int vis[100001];
struct node
{int pos;int step;
}q[10001];bool check(int x)
{if(x >= 0 && x <= 100000 && vis[x] == 0)return true;elsereturn false;
}int bfs()
{queue<node> q;node temp;temp.pos = n;temp.step = 0;vis[n] = 1;q.push(temp);while(!q.empty()){node temp2;temp = q.front();q.pop();temp2.step = temp.step + 1;temp2.pos = temp.pos - 1;if(check(temp2.pos)){if(temp2.pos == k)return temp2.step;q.push(temp2);vis[temp2.pos] = 1;}//cout << temp2.pos << " " << temp2.step << "\t";if(temp2.pos < k){temp2.pos = temp.pos + 1;if(check(temp2.pos)){if(temp2.pos == k)return temp2.step;q.push(temp2);vis[temp2.pos] = 1;}//cout << temp2.pos << " " << temp2.step << "\t";temp2.pos = 2 * temp.pos;if(check(temp2.pos)){if(temp2.pos == k)return temp2.step;q.push(temp2);vis[temp2.pos] = 1;}}//cout << temp2.pos << " " << temp2.step << endl;}
}int main()
{#ifdef DoubleQfreopen("in.txt","r",stdin);#endifwhile(~scanf("%d%d",&n,&k)){if(k == n){printf("0\n");continue;}memset(vis , 0 , sizeof(vis));printf("%d\n",bfs());}
}

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