搜索练习6/poj.org/problem3278 /Catch That Cow/简单的bfs模板

http://poj.org/problem?id=3278

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 89245		Accepted: 27980

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.

题意很简单：一维的坐标，从一点到另一点最少要走多少步；

思路：每一步有三种可能。+1，-1，*2；简单的bfs

AC代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<string>
#define LL long long
#define eps 1e-8
using namespace std;
const int mod = 1e7+7;
const int inf = 0x3f3f3f3f;
const int maxx = 100100;
const int N = 105;
int n,m;
int v[maxx];
struct node {int y,step;}start,endd;
int bfs(node x,node s)
{
    node p1,p2;
    queue<node>q;
    q.push(x);
    v[x.y]=1;
    while(!q.empty())
    {
       p1=q.front();
       if(p1.y==endd.y)
        return p1.step;
       q.pop();
       if(p1.y>0&&!v[p1.y-1])
       {
           p2=p1;
           p2.y-=1;
           p2.step++;
           v[p2.y]=1;
           q.push(p2);
       }
       if(p1.y<100000&&!v[p1.y+1])
       {
           p2=p1;
           p2.y+=1;
           v[p2.y]=1;
           p2.step++;
           q.push(p2);
       }
       if(p1.y*2<=100000&&!v[p1.y*2])
       {
           p2=p1;
           p2.y*=2;
           v[p2.y]=1;
           p2.step++;
           q.push(p2);
       }
    }
    return -1;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(v,0,sizeof(v));
        start.y=n,start.step=0;
        endd.y=m;
        printf("%d\n",bfs(start,endd));
    }
}