C - Catch That Cow POJ - 3278

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.

 

BFS+剪枝:

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
//#include <unordered_set>
//#include <unordered_map>
//#include <xfunctional>
using namespace std;
int dir[6][3] = { { 0,0,1 },{ 0,0,-1 },{ 0,-1,0 },{ 0,1,0 },{1,0,0},{-1,0,0} };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979;
#define ll  long long;
#define PII  pair<int, int>;
const int mod = 1e9 + 7;
const int marown = 1000 + 5;
//if(x<0 || x>=r || y<0 || y>=c)
struct node
{
    int pos,cnt;
};
int main()
{
    int n, k;
    cin >> n >> k;
    queue<node> que;
    vector<bool> visited(100001, false);
    if (k <= n)
    {
        cout << n - k;
    }
    else
    {
        node tmp;
        tmp.cnt = 0;
        tmp.pos = n;
        que.push(tmp);
        visited[n] = true;
        while (!que.empty())
        {
            node front = que.front();

            if (front.pos == k)
            {
                cout << front.cnt;
                break;
            }
            node tmpadd, tmpsub, tmp2;
            tmpadd.pos = front.pos + 1;
            tmpsub.pos = front.pos - 1;
            tmp2.pos = front.pos * 2;
            tmpadd.cnt = tmpsub.cnt = tmp2.cnt = front.cnt + 1;
            if (tmpadd.pos > 0 && tmpadd.pos <= 100000 && visited[tmpadd.pos]==false)
            {
                visited[tmpadd.pos] = true;
                que.push(tmpadd);
            }
            if (tmpsub.pos > 0 && tmpsub.pos <= 100000 && visited[tmpsub.pos] == false)
            {
                visited[tmpsub.pos] = true;
                que.push(tmpsub);
            }
            if (tmp2.pos > 0 && tmp2.pos <= 100000 && visited[tmp2.pos] == false)
            {
                visited[tmp2.pos] = true;
                que.push(tmp2);
            }
            que.pop();
        }
    }
    return 0;
}