catch that cow POJ 3278 搜索

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catch that cow POJ 3278 搜索

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>…

Catch That Cow POJ - 3278 [kuangbin带你飞]专题一简单搜索

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 m…

kuangbin专题专题一简单搜索 Catch That Cow POJ - 3278

题目链接:https://vjudge.net/problem/POJ-3278 题意:人可以左移动一格,右移动一格,或者移动到当前位置两倍下标的格子思路:把题意的三种情况跑bfs,第一个到达目的地的时间最短. #include <iostream> #include <string.h> #include<queue> #include <algorithm> using namespace std; #define rep(i,j,k) for(int…

（广搜）Catch That Cow -- poj -- 3278

链接: http://poj.org/problem?id=3278 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 62113 Accepted: 19441 Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a poi…

Catch That Cow POJ - 3278 bfs map超时，短路判断顺序。

题意:可以把n边为n+1,n-1,n*2问从n到k的最少变化次数. 坑:标题写了.有点不会写bfs了... ac代码 #define _CRT_SECURE_NO_WARNINGS #include<cstring> #include<cctype> #include<cstdlib> #include<cmath> #include<cstdio> #include<string> #include<stack> #in…

C - Catch That Cow POJ - 3278

//标准bfs #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <queue> using namespace std; ] = { , - }; struct node { int x, step; } s, ss; int bfs(int n, int k) { queue<node>q, qq; s.x =…

poj 3278 搜索

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 and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has t…