poj3278 Catch That Cow(简单的一维bfs)】的更多相关文章

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 DescriptionFarmer 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 l…
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…
Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 114140   Accepted: 35715 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…
链接:https://ac.nowcoder.com/acm/contest/984/L 来源:牛客网 Catch That Cow 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 32768K,其他语言65536K 64bit IO Format: %lld 题目描述 Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He sta…
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…
题目链接:http://poj.org/problem?id=3278 题目大意:给你两个数字n,k.可以对n执行操作(n+1,n-1,n*2),问最少需要几次操作使n变成k. 解题思路:bfs,每次走出三步n-1,n+1,n*2入队,直到最后找到答案为止.要注意: ①n不能变为负数,负数无意义,且无法用数组记录状态会runtime error ②n不用大于100000 代码: #include<cstdio> #include<cstring> #include<queue…
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 and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has…
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…
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): 6383    Accepted Submission(s): 2034 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: 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: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…
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: 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…
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: 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…
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…
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…
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…
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…
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 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…
BFS... -------------------------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<queue>   #define rep( i , n ) for( int i =…
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 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…
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 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…