3278 Catch That Cow BFS入门题

Catch That Cow
Time Limit: 2000MS      Memory Limit: 65536K
Total Submissions: 70410        Accepted: 22143
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
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#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
=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;
}

//!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
=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;
}