【BZOJ】1646: [Usaco2007 Open]Catch That Cow 抓住那只牛（bfs）

http://www.lydsy.com/JudgeOnline/problem.php?id=1646

这一题开始想到的是dfs啊，，但是本机测样例都已经re了。。。

那么考虑bfs。。。很巧妙?

首先我们得确定一个上下界。

当到达距离<0时显然不可能再走了（准确来说绝对不会比当前优），所以这里可以剪枝。

当到达距离>max(n, k)+1时也不能再走了（准确说不会比之前的优，比如说，你走到了k+2，但是之前就可在某个小于k的地方走×2的就走到了，那么就比这个少了1步）所以这里可以剪枝

然后bfs。。

#include <cstdio>

#include <cstring>

#include <cmath>

#include <string>

#include <iostream>

#include <algorithm>

#include <queue>

using namespace std;

#define rep(i, n) for(int i=0; i<(n); ++i)

#define for1(i,a,n) for(int i=(a);i<=(n);++i)

#define for2(i,a,n) for(int i=(a);i<(n);++i)

#define for3(i,a,n) for(int i=(a);i>=(n);--i)

#define for4(i,a,n) for(int i=(a);i>(n);--i)

#define CC(i,a) memset(i,a,sizeof(i))

#define read(a) a=getint()

#define print(a) printf("%d", a)

#define dbg(x) cout << #x << " = " << x << endl

#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }

inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }

inline const int max(const int &a, const int &b) { return a>b?a:b; }

inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int N=100005;

int n, k, front, tail, q[N], d[N], vis[N];

void add(int x) { if(vis[x]) return; vis[x]=1; q[tail++]=x; if(tail==N) tail=0; }

int main() {

	read(n); read(k);

	int mx=max(n, k)+1, t;

	q[tail++]=n;

	CC(d, 0x7f); d[n]=0;

	while(front!=tail) {

		t=q[front++]; if(front==N) front=0; vis[t]=0;

		if(t==k) break;

		int dt=d[t]+1;

		if(t>0 && dt+1<d[t-1]) d[t-1]=dt, add(t-1);

		if(t<mx && dt<d[t+1]) d[t+1]=dt, add(t+1);

		if((t<<1)<=mx && dt<d[t<<1]) d[t<<1]=dt, add(t<<1);

	}

	print(d[k]);

	return 0;

}

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?

农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来．

他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有

两种办法移动，步行和瞬移：步行每秒种可以让约翰从z处走到x+l或x-l处；而瞬移则可让他在1秒内从x处消失，在2x处出现．然而那只逃逸的奶牛，悲剧地没有发现自己的处境多么糟糕，正站在那儿一动不动．

那么，约翰需要多少时间抓住那只牛呢？

Input

* Line 1: Two space-separated integers: N and K

仅有两个整数N和K.

Output

* Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

最短的时间．

Sample Input

5 17
Farmer John starts at point 5 and the fugitive cow is at point 17.

Sample Output

OUTPUT DETAILS:

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.

HINT

Source

Silver