poj Catch That Cow （Bfs）

Catch That Cow

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 25   Accepted Submission(s) : 9

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
   
   
   
   

 

Source

PKU

入门题，注意边界和判重

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;

struct Node{
    int x,c;
}pos[200005];
queue <Node> q;
int n,k;
bool vist[200005];

int Bfs(){
    int p,countn;

    while(!q.empty()){
        p = q.front().x;
        countn = q.front().c;
        q.pop();
        if(p == k)
            return countn;
        if(!vist[p - 1] && p > 0){
            vist[p - 1] = 1;
            pos[p - 1].c = countn + 1;
            pos[p - 1].x = p - 1;
            q.push(pos[p - 1]);
        }
        if(p < k){
            if(!vist[p + 1]){
                vist[p + 1] = 1;
                pos[p + 1].c = countn + 1;
                pos[p + 1].x = p + 1;
                q.push(pos[p + 1]);
            }
            if(!vist[p * 2]){
                vist[p * 2] = 1;
                pos[p * 2].c = countn + 1;
                pos[p * 2].x = p * 2;
                q.push(pos[p * 2]);
            }
        }
    }
}

int main(){
    int i,j;

    while(scanf("%d%d",&n,&k) != EOF){
        memset(vist,0,sizeof(vist));
        memset(pos,0,sizeof(pos));
        while(!q.empty())//用到队列要记得初始化
            q.pop();
        vist[n] = 1;
        pos[n].x = n;
        pos[n].c = 0;
        q.push(pos[n]);
        printf("%d\n",Bfs());
    }
    return 0;
}