POJ 3278:Catch That Cow】的更多相关文章

Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 44613   Accepted: 13946 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: 2000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u Submit Status 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 ≤ 1…
The merchant Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 6864   Accepted: 2375 Description There are N cities in a country, and there is one and only one simple path between each pair of cities. A merchant has chosen some paths and w…
BFS算法与树的层次遍历很像,具有明显的层次性,一般都是使用队列来实现的!!! 常用步骤: 1.设置访问标记int visited[N],要覆盖所有的可能访问数据个数,这里设置成int而不是bool,基于一个考虑,多次循环时不用每次都清空visited,传递进去每次一个数字即可,比如第一次标记为1,判断也采用==1,之后递加即可. 2.设置一个node,用来记录相关参数和当前的步数,比如: struct node { int i; int j; int k; int s;//步数 }; 3.设计…
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 /* BFS简单题:考虑x-1,x+1,x*2三种情况,bfs队列练练手 */ #include <cstdio> #include <iostream> #include <algorithm> #include <map> #include <queue> #include <set> #include <cmath> #include <cstring> using namespace std…
Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 88732   Accepted: 27795 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…
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…
题目链接:http://poj.org/problem?id=3278 Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 124528   Accepted: 38768 Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. H…
catch that cow POJ 3278 搜索 题意 原题链接 john想要抓到那只牛,John和牛的位置在数轴上表示为n和k,john有三种移动方式:1. 向前移动一个单位,2. 向后移动一个单位,3. 移动到当前位置的二倍处.输出移动的最少次数. 解题思路 使用搜索,准确地说是广搜,要记得到达的位置要进行标记,还有就是减枝. 详情见代码实现. 代码实现 #include<cstdio> #include<cstring> #include<iostream>…
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 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: 2000MS   Memory Limit: 65536K Total Submissions: 58072   Accepted: 18061 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…
第一篇博客,格式惨不忍睹.首先感谢一下鼓励我写博客的大佬@Titordong其次就是感谢一群大佬激励我不断前行@Chunibyo@Tiancfq因为室友tanty强烈要求出现,附上他的名字. Catch That Cow(POJ3278) BFS入门题,然鹅我还是WA了四五发,因为没注意,位置0是可以访问的.再者就是初始位置在push之后,要标记为已经访问. 图片挺不错,我们地大(武汉)的旖旎风光,放松一下. 题目链接:POJ3278 Description Farmer John has be…
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…
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…
Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 46715   Accepted: 14673 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: 2000MS   Memory Limit: 65536K Total Submissions: 61826   Accepted: 19329 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 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 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…
Catch That Cow Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status 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 ≤…
POJ 3278 Catch That Cow 题目:你要去抓一头牛,给出你所在的坐标和牛所在的坐标,移动方式有两种:要么前一步或者后一步,要么移动到现在所在坐标的两倍,两种方式都要花费一分钟,问你最小花费时间恰好到达牛所在的地方. 思路:BFS求最优解,移动有三种情况,前后,和移动两倍位置,不过注意的地方是,当牛的坐标比你小,你只能一步步往后倒退,这个需要特判. #include<cstdio> #include<cmath> #include<cstring> #i…
题目链接:http://poj.org/problem?id=3278 Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 97563   Accepted: 30638 Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He…
Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 73973   Accepted: 23308 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…
http://poj.org/problem?id=3278                                                                                  Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 47010   Accepted: 14766 Description Farmer John has been infor…
Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 87152   Accepted: 27344 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 star…
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…
Catch That Cow Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8616    Accepted Submission(s): 2714 Problem Description Farmer John has been informed of the location of a fugitive cow and wants…
HDU 2717 题目大意:在x坐标上,农夫在n,牛在k.农夫每次可以移动到n-1, n+1, n*2的点.求最少到达k的步数. 思路:从起点开始,分别按x-1,x+1,2*x三个方向进行BFS,最先找到的一定是最小的步数. /* HDU 2717 Catch That Cow --- BFS */ #include <cstdio> #include <cstring> #include <queue> using namespace std; ]; int n, k…
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…