POJ 3278 Catch That Cow

Catch That Cow


Time Limit: 2000MS Memory Limit: 65536K

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.

Source

USACO 2007 Open Silver

题意：

给定两个整数n和k有n+1、n-1 、n*2， 3种操作使得n==k

输出最少的操作次数

关键在剪枝

#include <iostream>#include <cstdio>#include <queue>#include <cstring>using namespace std;int N , K ;int ans[100001] ;                        //记录步数void bfs( ) {    queue <int> Q ;    while(!Q.empty()) {        Q.pop() ;    }    Q.push(N) ;    while(!Q.empty()) {       int init = Q.front()  ;        Q.pop() ;        // +1        int temp = init + 1 ;        if( temp == K ){ if(ans[temp])            ans[temp] = min(ans[init] + 1 , ans[temp]) ;            else ans[temp] = ans[init] + 1 ;                return ;}        else if( temp <= 100000 && temp < K ) {  // 步数小于K才加            if(!ans[temp]){             ans[temp] = ans[init] + 1 ;            Q.push(temp) ;           }        }        //-1        temp = init -1 ;        if( temp == K ) {            if(ans[temp])            ans[temp] = min(ans[init] + 1 , ans[temp]) ;            else ans[temp] = ans[init] + 1 ;                return ;}        else if( temp >= 0  ) {                 //注意大于0           if(!ans[temp]){             ans[temp] = ans[init] + 1 ;            Q.push(temp) ;           }        }        //乘2        temp = init*2 ;        if( temp == K  ){if(ans[temp])            ans[temp] = min(ans[init] + 1 , ans[temp]) ;            else ans[temp] = ans[init] + 1 ;                return ;}        else if( temp <= 100000 && temp < 2*K ) {  //剪枝，若>=2K 无意义            if(!ans[temp]){             ans[temp] = ans[init] + 1 ;            Q.push(temp) ;           }        }    }}int main() {    while(~scanf("%d%d",&N,&K)) {            memset(ans,0,sizeof(ans)) ;            if( N == K) {                printf("0\n") ;                continue ;            }        bfs() ;        printf("%d\n",ans[K]) ;    }    return 0 ;}