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                Problem 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.

 

對所有狀況進行一次搜尋，最先找到的肯定就是時間最少的。

#include
 
 <stdio.h>#include <string.h>#include <queue>using namespace std;const int N = 1000000;int map[N+10];int n,k;struct node{    int x,step;};int check(int x){    if(x<0 || x>=N || map[x])        return 0;    return
 
 1;}int bfs(int x){    int i;    queue<node> Q;    node a,next;    a.x = x;    a.step = 0;    map[x] = 1;    Q.push(a);    while(!Q.empty())    {        a = Q.front();        Q.pop();        if(a.x == k)            return a.step;        next = a;        //每次都將三種狀況加入佇列之中        next.x = a.x+1;        if(check(next.x))        {            next.step = a.step+1;            map[next.x] = 1;            Q.push(next);        }        next.x = a.x-1;        if(check(next.x))        {            next.step = a.step+1;            map[next.x] = 1;            Q.push(next);        }        next.x = a.x*2;        if(check(next.x))        {            next.step = a.step+1;            map[next.x] = 1;            Q.push(next);        }    }    return -1;}int main(){    int ans;    while(~scanf("%d%d",&n,&k))    {        memset(map,0,sizeof(map));        ans = bfs(n);        printf("%d\n",ans);    }    return 0;}

 

           

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