POJ-3278(BFS)

题目：

                                                                                                                                                         Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K

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


分析：题目大意是在数轴上给定任意起点n、终点k，对任意的点坐标x每次有3种走法：x-1，x+1，2*x。每走一次花费时间为1minute，问从n到k最少需要花费多少时间？

      该题是最短路径问题，于是可以用BFS搜索，每次往三个方向BFS，直到到达k，每走一步记录当前时间。

//simonPR
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=100005;
int n,k,vis[maxn];
struct catche{
    int site,times;      //记录点和所花的时间。
};
queue<catche> q;
void init();
void init(){
    while(!q.empty())
        q.pop();
    memset(vis,0,sizeof(vis));
}
int bfs(int n,int k);
int bfs(int n,int k){
    if(n==k) return 0;
    catche now,news,start;
    start.site=n;
    start.times=0;
    q.push(start);         //将起点入队列。
    vis[n]=1;
    while(!q.empty()){
        int ts;
        now=q.front();
        q.pop();
        for(int i=0;i<3;i++){      //分三个方向BFS。
            if(i==0) ts=now.site-1;
            else if(i==1) ts=now.site+1;
            else ts=2*now.site;
            if(ts>maxn||vis[ts]==1) continue;       //如果已经访问过了或超出数据范围则跳过。
            if(ts==k) return now.times+1;           //到达终点K，返回时间。
            if(vis[ts]==0)                          //更新点信息，并将新点入队列。
            {
                vis[ts]=1;
                news.site=ts;
                news.times=now.times+1;
                q.push(news);
            }
        }
    }
}
int main()
{
    scanf("%d%d",&n,&k);
    init();                    //初始化。
    int ans=bfs(n,k);
    printf("%d
",ans);
    return 0;
}