POJ - 3278 Catch That Cow

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分类专栏：广度搜索 bfs 文章标签： bfs

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Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 116403		Accepted: 36372

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

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

题目大意：农夫约翰被告知了一头逃亡牛的位置，并想立即赶上她。他开始于一个点N（0≤ N ≤100,000)和牛是在点K（0≤ K≤100,000），农民约翰有两种运输方式：步行和传送。

*散步：FJ可以从任何点移动X到点X - 1或X在单个分钟+ 1
*隐形传态：FJ可以从任何点移动X到点2× X在一分钟。

如果母牛不知道它的追求，根本不动，那么农夫约翰需要多长时间才能找回它？

解题思路：借助队列，进行广度搜索，最先跳出的是最短时间。

#include <iostream>
#include<stdio.h>
#include <string.h>
#include <queue>
using namespace std;
int ans[100005];
int vis[100005];
int N,K;
queue<int>q;
int bfs(int pos,int end){//运用队列进行bfs搜索
    q.push(pos);
    ans[pos]=0;//步数初始化，第0步
    vis[pos]=1;//对第一个数据标记
    int x=0;
    while (!q.empty()) {//对队列判断，是否循环
        int t=q.front();//记录第一个数据
        q.pop();//清空队列
        
        for (int i=0; i<3; i++) {//列举每一种情况
            if (i==0) x=t-1;
            if (i==1) x=t+1;
            if(i==2)  x=2*t;
            
            if (x>2*end||x<0||x>=100001) continue;//限制x的范围
            
            if (!vis[x]){//对每一种情况判断，是否走过，防止死循环
                vis[x]=1;//对走过的进行标记
                q.push(x);//进入下一步
                ans[x]=ans[t]+1;//记录时间
                if (x==end) return ans[x];//判断条件，是否跳出；
            }
            
        }
    }
    return ans[x];
}
int main() {
    while(scanf("%d%d",&N,&K)!=EOF) {
        memset(vis, 0, sizeof(vis));//初始化数组
        memset(ans, 0, sizeof(ans));//初始化数组
        while (!q.empty()) q.pop();//清空队列
        if (N>=K){
            printf("%d",N-K);//如果n大于k只能减
        }else printf("%d",bfs(N, K));//另一种情况
        
    }
    return 0;
}