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HDU2717-Catch That Cow （BFS入门）
题目传送门：http://acm.hdu.edu.cn/showproblem.php?pid=2717
Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14880 Accepted Submission(s): 4495
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?
* 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.题意：
从a走到b，你可以向前走一步，向后走一步或者走a*2步，求最少多少步
1 #include<stdio.h> 2 #include<string.h> 3 #include<queue> 4 using namespace std; 5 int v[100010],t[100010]; 6 int main() 7 { 8 int n,a,b,i,f; 9 while(scanf("%d%d",&a,&b)!=EOF)10 {11 memset(t,0,sizeof(t));12 memset(v,0,sizeof(v));13 queue<int > q;14 q.push(a);15 v[a]=1;16 if(a==b)17 {18 printf("0\n");19 continue;20 }21 while(!q.empty())22 {23 int ll=q.front(),r=ll-1,l=ll+1,z=ll*2;24 q.pop();25 //printf("ll=%d\n",ll);26 if(z>=0&&z<=100000&&!v[z])27 {28 q.push(z);29 v[z]=v[ll]+1;30 // printf("v[z]=%d z=%d\n",v[z],z);31 }32 if(l>=0&&l<=100000&&!v[l])33 {34 q.push(l);35 v[l]=v[ll]+1;36 // printf("v[l]=%d l=%d\n",v[l],l);37 }38 if(r>=0&&r<=100000&&!v[r])39 {40 q.push(r);41 v[r]=v[ll]+1;42 // printf("v[r]=%d r=%d\n",v[r],r);43 }44 45 if(l==b||r==b||z==b)46 {47 48 break;49 }50 51 }52 53 printf("%d\n",v[b]-1);54 55 }56 return 0;57 58 }
HDU2717-Catch That Cow （BFS入门）
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