B-POJ-3278 Catch That Cow

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.

经典的BFS题目，每次3种选择

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;
int flag[100001];
void BFS(int N,int K){
queue<int> a;
int b;
a.push(N);
while(!a.empty()){
b=a.front();
a.pop();
if(b==K){
printf("%d",flag[b]);
return;
}
if(b+1<=100000&&flag[b+1]==0){
flag[b+1]=flag[b]+1;
a.push(b+1);
}
if(b-1>=0&&flag[b-1]==0){
flag[b-1]=flag[b]+1;
a.push(b-1);
}
if(2*b<=100000&&flag[2*b]==0){
flag[2*b]=flag[b]+1;
a.push(2*b);
}
if(b-1<0||2*b>100000||b+1>100000)
continue;
//开始将pop放在这里导致进死循环。。
}
return;
}
int main(){
int N,K;
memset(flag,0,sizeof(flag));
scanf("%d%d",&N,&K);
if(N<K)
BFS(N,K);
else
printf("%d",N-K);
return 0;
}

题目很简单，但开始没注意pop的位置，找了一晚上。。