BFS PKU 3278

一道很简单的，典型的BFS。

代码虐我千百遍，我待代码如初恋

Catch That Cow

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

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <ctype.h>
#include <limits.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <deque>
using namespace std;
#define MAXN 100000 + 10

int vis[MAXN];
int k;
struct node{
    int num;
    int sum;
    //int vis;
}a[MAXN];

void BFS(int n){
    queue<node>q;
    node front;
    node rear;
    front.num = n;
    vis[n] = 1;
    front.sum = 0;
    q.push(front);
    int mark1;
    int mark2;
    int mark3;
    //int mark4;
    //int sum = 0;
    while(!q.empty()){
        front = q.front();
        q.pop();
        if(front.num == k){
            printf("%d\n",front.sum);
            break;
        }
        mark1 = front.num - 1;
        mark2 = front.num + 1;
        mark3 = front.num*2;
        if(mark1>=0 && vis[mark1]==0){
            rear.num = mark1;
            rear.sum = front.sum+1;
            //sum++;
            vis[mark1] = 1;
            q.push(rear);
        }
        if(mark2<=100000 && vis[mark2]==0){
            rear.num = mark2;
            //sum++;
            rear.sum = front.sum + 1;
            vis[mark2] = 1;
            q.push(rear);
        }
        if(mark3<=100000 && vis[mark3]==0){
            rear.num = mark3;
            //sum++;
            rear.sum = front.sum + 1;
            vis[mark3] = 1;
            q.push(rear);
        }
    }
}

int main(){
    int n;
    while(~scanf("%d%d",&n,&k)){
        memset(vis,0,sizeof(vis));
        //memset(a,0,sizeof(a));
        BFS(n);
    }

    return 0;
}