ACM刷题之POJ———— Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 84879 Accepted: 26650

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.

一道bfs题，注意判断条件放在前面，要写vis 否则会死循环

下面是ac代码

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<algorithm>#include<map>#include<set>#include<queue>#include<string>#include<iostream>using namespace std;#define MID(x,y) ((x+y)>>1)#define CLR(arr,val) memset(arr,val,sizeof(arr))#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;const int N=2e5+7;typedef struct fi{int dis;int bu;}fi;fi fam,now,temps;queue<fi>q;int n,x,cnt;int vis[N];void bfs(){while(!q.empty()){now=q.front();q.pop();temps=now;if(now.dis==x){cnt=now.bu;break;}++now.bu;if(now.dis<x){if(!vis[temps.dis*2]){now.dis=temps.dis*2;q.push(now);vis[now.dis]=1;}if(!vis[temps.dis+1]){now.dis= temps.dis+1;q.push(now);vis[now.dis]=1;}if(!vis[temps.dis-1]){now.dis= temps.dis-1;q.push(now);vis[now.dis]=1;}}else {if(!vis[temps.dis-1]){now.dis= temps.dis-1;q.push(now);vis[now.dis]=1;}}}}int main(){//freopen("f:/input.txt", "r", stdin);int i,j;while(scanf("%d%d",&fam.dis,&x)!=EOF){if(fam.dis==x){printf("0\n");continue;}fam.bu=0;cnt=0;q.push(fam);CLR(vis,0);vis[fam.dis]=1;bfs();printf("%d\n",cnt);}}