POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 64397 Accepted: 20226

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

= =|| 训练队同队的队员AC之后一直在向我炫耀 估计也是做不成队友了（手动再见）

初学STL写的不太好，刚开始还忘记了n>k的情况

#include <cstdio>#include <cstdlib>#include <cstring>#include <queue>using namespace std;bool vis[200001];typedef struct{    int data,step;}position;position father,current;queue<position> bus;int vplus,vreduce,vadd;int main(void){    int n,k;    scanf("%d %d",&n,&k);if(n==k){printf("0\n");return 0;}if(n>k){printf("%d\n",n-k);return 0;}    current.step=0;    current.data=n;    bus.push(current);    while(!bus.empty())    {        father=bus.front();        bus.pop();        if(father.data==k)        {printf("%d\n",father.step);return 0;        }        vis[father.data]=true;        vplus=father.data+father.data;        vreduce=father.data-1;        vadd=father.data+1;        current.step=father.step+1;        if(vplus<200001&&!vis[vplus])        {            current.data=vplus;            bus.push(current);        }        if(vadd<200001&&!vis[vadd])        {            current.data=vadd;            bus.push(current);        }        if(vreduce>-1&&!vis[vreduce])        {            current.data=vreduce;            bus.push(current);        }    }    return (0);}

下面是在其他地方找到的一份代码，也AC了不过时间好短w

#include<iostream>#include<queue>using namespace std;#define M 100001#define INF 10000000struct POINT{int pos;int step;}now,next;queue<POINT>Q;bool visited[M];int n,k;int bfs(){while(!Q.empty())Q.pop();now.pos=n;now.step=0;visited[now.pos]=true;Q.push(now);while(!Q.empty()){now=Q.front();Q.pop();next=now;for(int i=0;i<3;i++){if(i==0) next.pos=now.pos+1;if(i==1)next.pos=now.pos-1;if(i==2)next.pos=now.pos*2;next.step=now.step+1;if(next.pos==k)return next.step;if(next.pos<0||next.pos>M)continue;if(!visited[next.pos]){visited[next.pos]=true;Q.push(next);}}}return INF;}int main(){while(cin>>n>>k){memset(visited,false,sizeof(visited));if(n<k)cout<<bfs()<<endl;if(n==k)cout<<0<<endl;if(n>k)cout<<n-k<<endl;}}

>_< 继续学习www