牛客假日团队赛5 L Catch That Cow HDU 2717 （BFS）

链接：https://ac.nowcoder.com/acm/contest/984/L
来源：牛客网

Catch That Cow
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
题目描述
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?
输入描述:
Line 1: Two space-separated integers: N and K
输出描述:
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
示例1
输入
复制
5 17
输出
复制
4
说明
Farmer John starts at point 5 and the fugitive cow is at point 17.
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.

题意：
给你一个位置n，一个位置k，都在水平的坐标轴上，你每一次可以向左或者向右走一步，或者从当先的x移动到2*x 位置，问你从n到k最小需要多少步？
思路：
直接BFS搜，用一个数组vis 维护 每一个位置是否被访问，由于BFS的性质，一个位置如果已经被访问过了，就不需要再访问了，因为当前的步数一定大于等于访问时候的步数。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 2000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
bool vis[maxn];
queue<pii> q;
int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    int n,k;
    cin>>n>>k;
    vis[n]=1;
    q.push(mp(n,0));
    pii t;
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        if(t.fi==k)
        {
            cout<<t.se<<endl;
            break;
        }else
        {
            if(t.fi+1<maxn&&!vis[t.fi+1])
            {
                vis[t.fi+1]=1;
                q.push(mp(t.fi+1,t.se+1));
            }
            if(t.fi-1>=0&&!vis[t.fi-1])
            {
                vis[t.fi-1]=1;
                q.push(mp(t.fi-1,t.se+1));
            }
            if(t.fi*2<maxn&&!vis[t.fi*2])
            {
                vis[t.fi*2]=1;
                q.push(mp(t.fi*2,t.se+1));
            }
        }
    }
    return 0;
}
 
inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}