poj3278 Catch That Cow

 

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 73973		Accepted: 23308

 

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

QAQ，写循环队列wa了好久。。。默默改大队列范围。

可行性剪枝：第一，当前点在牛的左边才进行右移；第二，当前点不能为负数。

15793310

ksq2013

3278

Accepted

1816K

32MS

G++

1159B

2016-07-23 19:37:16

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int n,cow;
bool vis[200100];
struct que{
    int fmr,stp;
}q[201000];
int bfs()
{
    int head=0,tail=1;
    q[0].fmr=n;
    q[0].stp=0;
    vis[n]=1;
    while(head^tail){
        que now=q[head++];
        //if(head==10000)head=0;
        if(!(now.fmr^cow))return now.stp;
        if(now.fmr-1>=0&&!vis[now.fmr-1]){
            vis[now.fmr-1]=1;
            q[tail].fmr=now.fmr-1;
            q[tail].stp=now.stp+1;
            tail++;
            //if(tail==10000)tail=0;
        }
        if(now.fmr<=cow&&!vis[now.fmr+1]){
            vis[now.fmr+1]=1;
            q[tail].fmr=now.fmr+1;
            q[tail].stp=now.stp+1;
            tail++;
            //if(tail==10000)tail=0;
        }
        if(now.fmr<=cow&&!vis[now.fmr<<1]){
            vis[now.fmr<<1]=1;
            q[tail].fmr=now.fmr<<1;
            q[tail].stp=now.stp+1;
            tail++;
            //if(tail==10000)tail=0;
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&cow)){
        memset(q,0,sizeof(que));
        memset(vis,0,sizeof(vis));
        printf("%d\n",bfs());
    }
    return 0;
}