catch the cow (P3278)

一开始就用我习惯的queue的方法进行广搜，结果就后面重复的枝条太多了。

后来用状态记录就OK

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<sstream>#include<algorithm>#include<queue>using namespace std;struct my{int num,t;};int ans;int s,e;my go[300000];bool vist[300000];void bfs(){memset(vist,false,sizeof(vist));int head,tail;head=tail=0;vist[s]=true;go[head].num=s;go[head].t=0;my cur;while (head<=tail){my d=go[head++];if (d.num==e){ans=d.t;return;}if (d.num-1>=0 && !vist[d.num-1]){tail++;go[tail].num=d.num-1;go[tail].t=d.t+1;vist[d.num-1]=true;}if (d.num<=e && !vist[d.num+1]){tail++;go[tail].num=d.num+1;go[tail].t=d.t+1;vist[d.num+1]=true;}if (d.num<=e && !vist[d.num*2] && d.num){tail++;go[tail].num=d.num*2;go[tail].t=d.t+1;vist[d.num*2]=true;}}}int main(){int i,j,k;while (cin>>s>>e){bfs();cout<<ans<<endl;}return 0;}

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 34076 Accepted: 10496

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX + 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 andK

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.

Source