POJ 3728 Catch That Cow (广搜)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 42564		Accepted: 13225

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

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.

#include<cstdio>
#include<string.h>
#include<iostream>
#include<queue>
#include<algorithm>using namespace std;int vis[100010];
int n,k;struct stu
{int ans;int sum;
};int bfs()
{queue <stu> q;
//    while(q.size())
//        q.pop();memset(vis,0,sizeof(vis));stu kaishi={n,0};q.push(kaishi);vis[kaishi.ans]=1;while(q.size()){stu xian=q.front();stu xia;q.pop();int i;for(i=0;i<3;i++){if(i==0) xia.ans=xian.ans+1;if(i==1) xia.ans=xian.ans-1;if(i==2) xia.ans=xian.ans*2;xia.sum=xian.sum+1;if(xia.ans==k)return xia.sum;if(xia.ans>=0&&xia.ans<=100010&&!vis[xia.ans]){vis[xia.ans]=1;q.push(xia);}}}return 0;
}int main()
{while(scanf("%d%d",&n,&k)!=EOF){if(n>=k){printf("%d\n",n-k);}elseprintf("%d\n",bfs());}return 0;
}

POJ 3728 Catch That Cow (广搜)相关推荐

poj 3278 Catch That Cow 广搜
hdu 2717 Catch That Cow,题目链接 Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Jav ...
POJ 3278 / hdu 2717 Catch That Cow (广搜）
POJ 3278 HDU 2717 广搜题,用一个数组标记就可以过,不标记的话会超内存. 另外,poj的数据要比hdu强一些,比如0 100,这种数据.不特判的话会RE.不过如果不特判,在poj上用C ...
Catch That Cow——广搜
Catch That Cow Problem Description Farmer John has been informed of the location of a fugitive cow a ...
poj 3728 Catch That Cow （[kuangbin带你飞]专题一简单搜索）
题目大意:题目链接就是给你N,K,每次有三种惭怍+1,-1,*2,,问多少次操作能到K 解题思路,搜索直接算,.,,,哎,啥时候这种垃圾搜索我能直接A 啊,太菜了 #include<cstdi ...
POJ 3278 Catch That Cow
Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 30924 Accepted: 9536 D ...
BFS POJ 3278 Catch That Cow
题目传送门 1 /* 2 BFS简单题:考虑x-1,x+1,x*2三种情况,bfs队列练练手 3 */ 4 #include <cstdio> 5 #include <iostrea ...
poj 3278 Catch That Cow（广搜）
Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 45087 Accepted: 14116 ...
POJ - 3278 Catch That Cow 简单搜索
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. ...
POJ 3278 Catch That Cow BFS
Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 32071 Accepted: 9866 D ...

POJ 3728 Catch That Cow (广搜)

POJ 3728 Catch That Cow (广搜)相关推荐

最新文章

热门文章