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

 

posted @ 2015-06-07 12:41  JSZX11556  阅读(465)  评论(0编辑  收藏  举报