题目链接: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): 12615    Accepted Submission(s): 3902 Problem Description Farmer John has been…

传送门 Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 80273   Accepted: 25290 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 ≤ 10…

Catch That Cow 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…

Catch That Cow Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10166    Accepted Submission(s): 3179 Problem Description Farmer John has been informed of the location of a fugitive cow and wants…

Catch That Cow Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 67   Accepted Submission(s) : 22 Problem Description Farmer John has been informed of the location of a fugitive cow and wants to c…

Catch That Cow bfs代码 #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<string> #include<vector> #include<stack> #include<bitset> #include<cstdlib> #include<cmath>…

Catch That Cow Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6383    Accepted Submission(s): 2034 Problem Description Farmer John has been informed of the location of a fugitive cow and wants…

Catch That Cow Descriptions: 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 numbe…

Catch That Cow Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9753    Accepted Submission(s): 3054 Problem Description Farmer John has been informed of the location of a fugitive cow and wants…

本文来源于:http://blog.csdn.net/svitter 意甲冠军:给你一个数字n, 一个数字k.分别代表主人的位置和奶牛的位置,主任能够移动的方案有x+1, x-1, 2*x.求主人找到奶牛的时间(奶牛不移动) 题解:最基础的BFS可是脑子犯抽WA了3遍- = 注意: 1.数组范围1~1<<5 2.visit去重.(BFS最基础的) 代码: #include <iostream> #include <stdio.h> #include <string…

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. F…

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 Jo…

题目: 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 t…

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 Jo…

题目 #include<stdio.h> #include<string.h> #include<queue> #include<algorithm> using namespace std; struct tt { int step,tem; }; ]; queue <tt> q; int bfs(tt s,tt e) { tt front,temp; q.push(s); visit[s.tem]=; while(!q.empty()) {…

题目:http://poj.org/problem?id=3278 题意: 给定两个整数n和k 通过 n+1或n-1 或n*2 这3种操作,使得n==k 输出最少的操作次数 #include<stdio.h> #include<string.h> #include<queue> using namespace std; ]; struct node { int x,step; }; int bfs(int n,int k) { if(n==k) ; queue<n…

题目链接:http://poj.org/problem?id=3278 这几次都是每天的第一道题都挺顺利,然后第二道题一卡一天. = =,今天的这道题7点40就出来了,不知道第二道题在下午7点能不能出来.0 0 先说说这道题目,大意是有个农夫要抓牛,已知牛的坐标,和农夫位置. 并且农夫有三种移动方式,X + 1,X - 1,X * 2.问最少几步抓到牛. 開始觉得非常easy的,三方向的BFS就能顺利解决.然后在忘开标记的情况下直接广搜,果然TLE,在你计算出最少位置之前.牛早跑了. 然后反应过…

相比于POJ2251的三维BFS,这道题做法思路完全相同且过程更加简单,也不需要用结构体,check只要判断vis和左右边界的越界情况就OK. 记得清空队列,其他没什么好说的. #include<iostream> #include<queue> #include<cstring> #include<cstdio> using namespace std; const int maxn=100001; bool vis[maxn]; int step[max…

POJ 3278 Catch That Cow(赶牛行动) Time Limit: 1000MS    Memory Limit: 65536K Description - 题目描述 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…

Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 45648   Accepted: 14310 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 ≤ 100,00…

Catch That Cow Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9276    Accepted Submission(s): 2907 Problem Description Farmer John has been informed of the location of a fugitive cow and wants…

Catch That Cow 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…

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): 20259    Accepted Submission(s): 5926 Problem Description Farmer John has been i…

Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 109702   Accepted: 34255 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 ≤ 100,0…

题目链接:Catch That Cow 题目大意 FJ丢了一头牛,FJ在数轴上位置为n的点,牛在数轴上位置为k的点.FJ一分钟能进行以下三种操作:前进一个单位,后退一个单位,或者传送到坐标为当前位置两倍的地方.求FJ能找到牛的最短时间. 思路 BFS.在每一个点有三种选择,前进,后退,或者传送.要注意的是,由于有后退的过程,所以可能会造成环,导致队列长度很长就直接MLE了.因此要用一个vis数组来控制不能选择已经去过的地方. 题解 #include <iostream> #include &l…

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…

题目链接:http://poj.org/problem?id=3278 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…

转载请注明出处:https://blog.csdn.net/Mercury_Lc/article/details/82693928作者:Mercury_Lc 题目链接 题解:给你x.y,x可以加1.减1.或者变成2*x,问通过最少的次数来让x等于y,这是最基础的bfs,就是把x通过一次的+1.-1.*2得到的数都放到队列里面,再把这些通过一次操作得到的数进行相同的操作+1.-1.*2,因为用个结构体来存放这个数是第几次操作得到的,所以只要一旦发现这个数,一定是通过最小的次数得到的. #inclu…

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 Jo…

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. F…