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?
Output
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
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.
注意数组要开大一些,否则会Runtime Error
#include<stdio.h>
#include<stack>
#include<queue>
#include<string.h>
using namespace std;
int state[200010];
int N,K;
int dir[3][2]={{1,-1},{1,1},{2,0}};
queue<int> record;
int BFS(){
if(N==K) return 0;
record.push(N);
state[N]=1;
while(!record.empty()){
int cur,temp;
cur=record.front();
record.pop();
for(int a=0;a<3;a++){
temp=cur*dir[a][0]+dir[a][1];
if(temp==K) return state[cur];
if(temp>=0&&temp<200010){
if(!state[temp]){
record.push(temp);
state[temp]=state[cur]+1;
}
}
}
}
}
int main(){
memset(state,0,sizeof(state));
cin>>N>>K;
int step;
step=BFS();
cout<<step<<endl;
return 0;
}