POJ 3728 Catch That Cow (广搜)

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 42564 Accepted: 13225

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.

#include<cstdio>#include<string.h>#include<iostream>#include<queue>#include<algorithm>using namespace std;int vis[100010];int n,k;struct stu{    int ans;    int sum;};int bfs(){    queue <stu> q;//    while(q.size())//        q.pop();    memset(vis,0,sizeof(vis));    stu kaishi={n,0};    q.push(kaishi);    vis[kaishi.ans]=1;    while(q.size())    {        stu xian=q.front();        stu xia;        q.pop();        int i;        for(i=0;i<3;i++)        {            if(i==0) xia.ans=xian.ans+1;            if(i==1) xia.ans=xian.ans-1;            if(i==2) xia.ans=xian.ans*2;            xia.sum=xian.sum+1;            if(xia.ans==k)                return xia.sum;            if(xia.ans>=0&&xia.ans<=100010&&!vis[xia.ans])            {                vis[xia.ans]=1;                q.push(xia);            }        }    }    return 0;}int main(){    while(scanf("%d%d",&n,&k)!=EOF)    {        if(n>=k)        {            printf("%d\n",n-k);        }        else        printf("%d\n",bfs());    }    return 0;}