标签：des style io ar for strong sp div on
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 48036 | Accepted: 15057 |
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
Output
Sample Input
5 17
Sample Output
4
题意：输入两个数n，k。求从n到k最少走多少步。可以前进1后退1或者当前的位置*2；
#include <stdio.h> #include <string.h> #include <stdlib.h> #include <algorithm> using namespace std; struct node { int x;//当前位置 int ans;//走的步数 }q[1000010]; int vis[1000010];//标记变量，该点是否被访问； int jx[]={-1,1};//后退1或者前进1； struct node t,f; int n,k; void bfs() { int i; int s=0,e=0;//指针模拟队列。往队列加e++ 往队列里提出数s++ memset(vis,0,sizeof(vis)); t.x=n;//当前初始位置 vis[t.x]=1;//标记为1代表访问过； t.ans=0;//初始位置步数为0； q[e++]=t;//把当前步数加人队列 while(s<e)//当队列不为空 { t=q[s++];//提出 if(t.x==k)//如果该数正好等于目标位置直接输出步数 { printf("%d\n",t.ans); break; } for(i=0;i<3;i++)//i=0后退一步，i=1前进一步，i=2此时的位置*2； { if(i==2) { f.x=t.x*2; } else { f.x=t.x+jx[i]; } if(f.x>=0&&f.x<=100000&&!vis[f.x]) { f.ans=t.ans+1; q[e++]=f; vis[f.x]=1; } } } } int main() { while(~scanf("%d %d",&n,&k)) { bfs(); } return 0; }
标签：des style io ar for strong sp div on
原文地址：http://blog.csdn.net/u013486414/article/details/40262377
