Catch That Cow POJ - 3278 [bfs][最短路]

Catch That Cow

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

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.

#include <iostream>
#include <string>
#include <queue>

using namespace std;
int v[200000]={0};
queue<int> line;

bool judge(int a){ //判断是否可以拓展
if (v[a] == 0 && a >= 0 && a <= 100000 ) return true;
else return false;
}

int main() {
int n,m;
while ( cin >> n >> m) {
int min=100000;
if(n == m) {
cout << '0' << endl;
continue;
}
memset(v, 0, sizeof(v));
while ( !line.empty()) {
line.pop();
}

v
= 1;
line.push(n);
int time=0;
int k; //拓展层数
while ( !line.empty()) {
k = line.front();
//            cout << "k:" <<  k;
line.pop();
if( judge(k+1) ) {line.push(k+1); v[k+1] += v[k]+1;}
if( judge(k-1) ) {line.push(k-1);  v[k-1] += v[k]+1;}
if (judge(k*2))  {line.push(k*2);  v[k*2] += v[k]+1;}

}
cout << v[m]-1 <<endl;

}

}