poj 3278 Catch That Cow (bfs搜索）

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?

#include <cstdio>
#include <cstring>
const int maxn=200030;
typedef struct Queue //一个结构体，保存这个点的位置，和到这个点所用的时间
{
    int count,step;
}Queue;
Queue queue[maxn];//队列数组
int visit[maxn];//访问数组
void bfs(int n,int k)
{
    int front=0,rear=0;
    queue[rear].step=n;//把初始位置入队
    queue[rear++].count=0;
    visit[n]=1;//标记访问
    while(front<rear)
    {
        Queue q=queue[front++];
        if(q.step==k)//找到终点位置
        {
            printf("%d\n",q.count);
            break;
        }
        if(q.step-1>=0 && !visit[q.step-1])//注意剪枝的条件，一定要注意可以等于0
        {
            visit[q.step]=1;
            queue[rear].step=q.step-1;//向后寻找一位
            queue[rear++].count=q.count+1;//时间+1
        }
        if(q.step<=k && !visit[q.step+1])//剪枝
        {
            visit[q.step+1]=1;
            queue[rear].step=q.step+1;//向前寻找一位
            queue[rear++].count=q.count+1;
        }
        if(q.step<=k && !visit[q.step*2])//剪枝
        {
            visit[q.step*2]=1;//传送的情况
            queue[rear].step=q.step*2;
            queue[rear++].count=q.count+1;
        }
    }
}
int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(visit,0,sizeof(visit));
        bfs(n,k);
    }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=200030;
queue<int>q;
int Count[maxn];
bool visit[maxn];
void bfs(int n,int k)
{
   q.push(n);
   visit[n]=1;
   Count[n]=0;
   while(!q.empty())
   {
       int x=q.front();
       q.pop();
       if(x==k)
       {
           printf("%d\n",Count[x]);
           break;
       }
       if(x-1>=0 && !visit[x-1])
       {
           q.push(x-1);
           visit[x-1]=1;
           Count[x-1]=Count[x]+1;
       }
       if(x<=k && !visit[x+1])
       {
           q.push(x+1);
           visit[x+1]=1;
           Count[x+1]=Count[x]+1;
       }
        if(x<=k && !visit[2*x])
       {
           q.push(x*2);
           visit[x*2]=1;
           Count[x*2]=Count[x]+1;
       }
   }
}
int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(visit,0,sizeof(visit));
        bfs(n,k);
    }
    return 0;
}

#include <iostream>
#include <queue>
#define SIZE 100001

using namespace std;

queue<int> x;
bool visited[SIZE];
int step[SIZE];

int bfs(int n, int k)
{
int head, next;
//起始节点入队
x.push(n);
//标记n已访问 
visited[n] = true;
//起始步数为0 
step[n] = 0;
//队列非空时 
while (!x.empty())
{
//取出队头 
head = x.front();
//弹出队头 
x.pop();
//3个方向搜索 
for (int i = 0; i < 3; i++)
{
if (i == 0) next = head - 1;
else if (i == 1) next = head + 1;
else next = head * 2;
//越界就不考虑了 
if (next > SIZE || next < 0) continue;
//判重 
if (!visited[next])
{
//节点入队 
x.push(next);
//步数+1 
step[next] = step[head] + 1;
//标记节点已访问 
visited[next] = true;
}
//找到退出 
if (next == k) return step[next];
}
}
}

int main()
{
int n, k;
cin >> n >> k;
if (n >= k)
{
cout << n - k << endl;
}
else
{
cout << bfs(n, k) << endl;
}
return 0;
}