3278 Catch That Cow BFS入门题

Catch That CowTime Limit: 2000MS      Memory Limit: 65536KTotal Submissions: 70410        Accepted: 22143DescriptionFarmer 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 Input5 17Sample Output4HintThe 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.SourceUSACO 2007 Open Silver[Submit]   [Go Back]   [Status]   [Discuss]

#include <iostream>#include <cstring>#include <queue>using namespace std;bool Hash[400001];  //!判断该点是否走过int main(void){    int N,K;    while(cin>>N>>K)    {        memset(Hash,false,sizeof(Hash));//初始化Hash表        pair<int,int> p;                //第一个代表横坐标，第二个代表走的步子        p.first=N;p.second=0;           //初始化        Hash[N]=true;                   //原点标记已走        queue< pair<int,int> > bfs;       //创建队列        bfs.push(p);                    //把原点设为第一个队列检索项        while(!bfs.empty())//队列有元素可操作        {            p=bfs.front();//获取队首            if(p.first==K)            {//已经找到了该元素，输出路径长，并结束搜索                cout<<p.second<<endl;                break;            }            p.second++;   //移动次数+1            pair<int ,int> q;            if(p.first<K)               {//当前点在目标点的左侧             //!*2                q=p;                q.first*=2;                if(Hash[q.first]==false&&q.first)                {//这个点没访问过，则从这点开始搜索                    Hash[q.first]=true;                    bfs.push(q);//压入队列                }              //! +1                q=p;                q.first+=1;                if(Hash[q.first]==false&&q.first)                {//这个点没访问过，则从这点开始搜索                    Hash[q.first]=true;                    bfs.push(q);//压入队列                }            }            if(p.first>0)            {//任意情况下都可以通过减1步继续搜索                q=p;                q.first--;                if(Hash[q.first]==false)                {//这个点没访问过，则从这点开始搜索                    Hash[q.first]=true;                    bfs.push(q);//压入队列                }            }            bfs.pop();//队首用完出列        }    }    return 0;}

来源： http://poj.org/problem?id=3278

//!结构体做法#include <iostream>#include <queue>#include <cstring>using namespace std;typedef struct{int now,step;}point;bool Hash[400001];//判断该点是否遍历过int N,K;int Cal(void){    memset(Hash,false,sizeof(Hash));    point P;P.now = N;P.step = 0;//初始化    Hash[N]=true;               //原点标记已走    queue<point> bfs;bfs.push(P);    while(!bfs.empty())    {        P = bfs.front();//获取队首        if(P.now == K) return P.step;        P.step ++; //步长+1        point Q;        if(P.now<K) ///目标左侧        {            ///+1            Q=P;Q.now++;            if(!Hash[Q.now]&&Q.now){Hash[Q.now] = true; bfs.push(Q);}            ///*2            Q=P;Q.now*=2;            if(!Hash[Q.now]&&Q.now){Hash[Q.now] = true; bfs.push(Q);}        }        if(P.now>0)///任意情况都可以-1        {            Q=P;Q.now--;            if(!Hash[Q.now]){Hash[Q.now] = true; bfs.push(Q);}        }        bfs.pop();//队首用完出列    }    return 0;}int main(){    while(cin>>N>>K) cout<<Cal()<<endl;    return 0;}