[poj] Catch That Cow--bfs

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.

三个搜索方向 +1, -1, *2  用bfs搜索 数组要开到最大值的2倍 注意剪枝和边界问题

#include <iostream>
#include <stdio.h>
#include <queue>
#include <cstring>
using namespace std;

const int Max = 210000;
int s, e;
struct node
{
    int x, step;
}now, Next;
bool v[210000];
queue<node>q;

int bfs()
{
    now.x = s;
    now.step = 0;
    v[s] = 1;
    q.push(now);
    while (!q.empty()) {
        now = q.front();
        q.pop();
        if (now.x == e)
            return now.step;

        Next.x = now.x*2;
        Next.step = now.step+1;
        if (Next.x >= 0 && Next.x <= Max && !v[Next.x]){
            q.push(Next);
            v[Next.x] = 1;
        }

        Next.x = now.x-1;
        Next.step = now.step+1;
        if (Next.x >= 0 && Next.x <= Max && !v[Next.x]){
            q.push(Next);  //若先检查v[Next.x] 会出现数组越界v[-1]的情况
            v[Next.x] = 1;
        }

        Next.x = now.x+1;
        Next.step = now.step+1;
        if (Next.x >= 0 && Next.x <= Max && !v[Next.x]){
            q.push(Next);
            v[Next.x] = 1;
        }

    }
    return 0;
}

int main()
{
    //freopen("1.txt", "r", stdin);
    cin >> s >> e;
    memset(v, 0, sizeof(v));
    cout << bfs();

    return 0;
}