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

BFS...

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<queue>

 

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

#define clr( x , c ) memset( x , c , sizeof( x ) )

 

using namespace std;

 

const int maxn = 200000;

const int inf = 0x3f3f3f3f;

 

queue< int > Q;

 

int d[ maxn ];

int main() {

// freopen( "test.in" , "r" , stdin );

int n , k;

cin >> n >> k;

clr( d , inf );

d[ n ] = 0;

Q.push( n );

while( ! Q.empty() ) {

int x = Q.front();

Q.pop();

if( x == k )

   break;

#define ok( x ) ( 0 <= x && x <= 100000 )

if( ok( x + 1 ) && d[ x + 1 ] > d[ x ] + 1 ) {

d[ x + 1 ] = d[ x ] + 1;

   Q.push( x + 1 );

   

}

   

if( ok( x - 1 ) && d[ x - 1 ] > d[ x ] + 1 ) {

d[ x - 1 ] = d[ x ] + 1;

   Q.push( x - 1 );

   

}

   

if( ok( x << 1 ) && d[ x << 1 ] > d[ x ] + 1 ) {

d[ x << 1 ] = d[ x ] + 1;

   Q.push( x << 1 );

   

}

   

}

cout << d[ k ] << "\n";

return 0;

}

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1646: [Usaco2007 Open]Catch That Cow 抓住那只牛

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 764  Solved: 361
[Submit][Status][Discuss]

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

4

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