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[n] = 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;

    }


}