poj3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal 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 pointN (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 orX + 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 andK

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 ksq20133278Accepted1816K32MSG++1159B2016-07-23 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;}