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
Output
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.
大致题意：FJ要去抓奶牛，他在n点，奶牛在k点，FJ每次可以选择：
1. 前进一步 花费1分钟；
2. 后退一步 花费1分钟；
3.前进当前点X的2倍步，即2×X，花费1分钟；
问最少需要多少分钟他能抓到奶牛。
典型搜索题，但是数据量太大，不能用深搜，只能用广搜。
搜索三个步骤。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#define N 100005
using namespace std;
typedef struct Que{
int num;
int step;
}que;
bool book[N];
queue<que> q;
void clear(queue<que> queue){
while(!queue.empty()) queue.pop();
}
int main(int argc, char *argv[]) {
int n,k,flag,ans;
cin>>n>>k;
//while(cin>>n>>k){
clear(q);
ans = 0;
flag = 0;
memset(book,0,sizeof(book));
if(n >= k) cout<<n-k<<endl;
else{
book[n] = true;
que tmpn;
tmpn.num = n;
tmpn.step = 0;
q.push(tmpn);
while(!q.empty()){
for(int cao = 1; cao <= 3; ++cao){
que tn;
switch(cao){
case 1:
tn.num = q.front().num + 1;
break;
case 2:
tn.num = q.front().num - 1;
break;
case 3:
tn.num = q.front().num * 2;
break;
}
if(tn.num > N || tn.num < 0 || book[tn.num]) continue;
if(!book[tn.num]){
tn.step = q.front().step + 1;
q.push(tn);
book[tn.num] = true;
}
if(q.back().num == k){
ans = q.back().step;
flag = 1;
break;
}
}
if(flag) break;
q.pop();
}
cout<<ans<<endl;
}
//}
return 0;
}