HDU 2727 Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5457    Accepted Submission(s): 1740

Problem 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

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.

 

Source

USACO 2007 Open Silver

 

Recommend

teddy

 

思路：BFS + 剪枝，我是一坑货，剪枝只会皮毛

 

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
using namespace std;
int n,k;
int map[100010];
struct Node
{
    int x;
    int step;
    bool operator < (const Node &t) const
     {
           return t.step < step;
     }
}info;
int BFS()
{
    priority_queue < Node > q;
    info.x = n;
    info.step = 0;
    q.push(info);
    map[info.x] = 1;
    while(!q.empty())
    {
        Node boss;
        boss = q.top();
        q.pop();
        if(boss.x == k)
            return boss.step;
        int a = boss.x;
        int step = boss.step;
        //printf("%d              %d\n",a,step);
        if(a * 2 <= 100000 && a < k && map[a * 2] == 0)
        {
                info.x = a * 2;
                info.step = step + 1;
                q.push(info);
                map[info.x] = 1;
                //printf("%d   %d\n",info.x,info.step);
                if(info.x == k)
                    return info.step;
        }
        if(a - 1 >= 0 && map[a - 1] == 0)
        {
                info.x = a - 1;
                info.step = step + 1;
                q.push(info);
                map[info.x] = 1;
                //printf("%d   %d\n",info.x,info.step);
                if(info.x == k)
                    return info.step;
        }
        if(a + 1 <= 100000 && map[a + 1] == 0)
        {
                info.x = a + 1;
                info.step = step + 1;
                q.push(info);
                map[info.x] = 1;
                //printf("%d   %d\n",info.x,info.step);
                if(info.x == k)
                    return info.step;
        }
    }
}
int main()
{
        while(~scanf("%d%d",&n,&k))
        {
              memset(map,0,sizeof(map));
              printf("%d\n",BFS());
        }
}