HDOJ2717 BFS水题

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16157    Accepted Submission(s): 4837

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

 

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好像没什么难度，标准的BFS过了。DFS应该也可以的。

#include <iostream>
#include <cstdio>
#include <ctime>
#include <cstring>
#include <queue>
using namespace std;

const int maxn = 1e5+10;
int n,k;
bool used[maxn<<1];
struct node {
   int pos,time;
}p,q;
queue<node>myqueue;

int bfs(){
    p.pos = n;
    p.time = 0;
    while (!myqueue.empty()) myqueue.pop();
    myqueue.push(p);
    while (!myqueue.empty()) {
        p = myqueue.front();
        myqueue.pop();
        if (p.pos == k) return p.time;
        if (p.pos<k){

            q.pos = p.pos-1;
            q.time = p.time+1;
            if (q.pos>=0 ) {
                if (!used[q.pos]) { //该位置之前还没标记过
                    used[q.pos] = 1;
                    myqueue.push(q);
                }
            }

            q.pos = p.pos+1;
            q.time = p.time+1;
            if (!used[q.pos]) { //该位置之前还没标记过
                used[q.pos] = 1;
                myqueue.push(q);
            }


            q.pos = p.pos<<1;
            q.time = p.time+1;
            if (!used[q.pos]) { //该位置之前还没标记过
                used[q.pos] = 1;
                myqueue.push(q);
            }
        }
        else {
            q.pos = p.pos-1;
            q.time = p.time+1;
            if (!used[q.pos]) { //该位置之前还没标记过
                used[q.pos] = 1;
                myqueue.push(q);
            }
        }
    }
}

int main(){
    while (cin >> n >> k){
       memset(used,0,sizeof(used));
       cout << bfs() << endl;
    }
    return 0;
}