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>…