Poj 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 58054		Accepted: 18053

DescriptionFarmer 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?InputLine 1: Two space-separated integers: N and KOutputLine 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

#include <iostream>#include<cstring>#include<cstdio>#include<queue>using namespace std;int n,m;struct node{    int x,ans;} ;int vis[1000010],mi;int jx[]={-1,1};void bfs(){    queue<node>q;    memset(vis,0,sizeof(vis));    node f1,f2;    int i,j,a;    f1.x=n,f1.ans=0;    q.push(f1);    mi=0x3f3f3f3f;    while(!q.empty())    {        f1=q.front();        q.pop();        if(f1.x==m)        {            if(mi>f1.ans)            {                mi=f1.ans;                continue;            }        }        if(f1.ans>=mi)            continue;        for(i=0;i<3;i++)        {            f2=f1;            if(i==2)                 f2.x=f2.x*2;//这里与i==1||i==0时都要改变f2.x的值，而不是设一个新值，因为还要把此数压入栈内            else                f2.x=f2.x+jx[i];            if(f2.x>=0&&f2.x<=100000&&vis[f2.x]==0)            {                vis[f2.x]=1;                f2.ans++;                q.push(f2);            }        }    }}int main(){    ios::sync_with_stdio(false);//注意写上加速语句防TLE    while(cin>>n>>m)    {        if(n==m)            printf("0\n");        else        {            bfs();            cout<<mi<<endl;        }    }    return 0;}