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…

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 //广搜,a是队列,step记录步数,vis记录哪些数被搜到过 #include<cstdio> #include<iostream> #define M 1000010 using namespace std; int step[M],a[M],vis[M]; int main() { ,tail=; scanf("%d%d",&n,&m); a[…

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 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…

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…

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…

题目链接 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 li…

个人心得:其实有关搜素或者地图啥的都可以用广搜,但要注意标志物不然会变得很复杂,想这题,忘记了标志,结果内存超时: 将每个动作扔入队列,但要注意如何更简便,更节省时间,空间 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 t…

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…