poj 3278 Catch That Cow (BFS) 题意: 牛逃跑了,主人用最短的时间追到牛(在直线上) 很裸的bfs ,郁闷啊忘了判断重复了,RE ,WA ! #include<stdio.h>#define N 100005int n,k;int stack[N],vis[N];int main(){ scanf("%d%d",&n,&k); int di=0,top=1,first=0,last=1; int sum=0;stack...
}intmain() {inti,j,k;while(scanf("%d%d",&n,&m)!=EOF) { memset(num,0,sizeof(num)); printf("%d\n",bfs()); }return0; }
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN(0 ≤N≤ 100,000) on a number line and the cow is at a pointK(0 ≤K≤ 100,000) on the same number line. Farmer John has two modes of transportation: walkin...
解析:直接bfs #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> using namespace std; typedef long long ll; const int maxn = 1e6+100; int n,k; int vis[maxn]; struct node { int x,step; node() {} node(int _x,int ...
POJ3278——Catch That Cow Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN(0 ≤N≤ 100,000) on a number line and the cow is at a pointK(0 ≤K≤ 100,000) on the same number line. Farmer John ...
POJ3278—Catch That Cow 题目描述 农夫约翰已被告知逃亡牛的位置,并希望立即抓住她。他开始于一个点N(0≤ N≤100,000)和牛是在点K (0≤ K≤100,000),他们在同一条线上。农夫约翰有两种交通方式:步行和乘车。 行走:约翰可以在一分钟内从任意点X移动到X_牛客网_牛客
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 modes of ...
POJ 3278 解题思路: 广度优先搜索(BFS)。一开始想使用深度优先搜索(DFS)来着,但是如果深搜的话会一直顺着某一个方式搜下去,不设计很多剪枝策略会很费时。我使用了白书上的BFS模板,全当复习广搜了。这个题需要注意的地方还有数组越界问题,因为有一个vis数组保存之前访问过的节点,每次放入新节点之前要保证节点没有...
POJ 3278 Catch That Cow(BFS) 题目链接:https://cn.vjudge.net/problem/POJ-3278 题意:给你 n 和 k ,可以对 n 进行 n+1、n-1和n*2的操作,问至少经过几次操作数字 n 能够到达 k 。 思路:从n开始BFS,每次将三种操作加入队列搜索,最先搜索到的肯定是操作最少的,由此对已经搜索过的点进行标记剪枝...
Catch That Cow Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN(0 ≤N≤ 100,000) on a number line and the cow is at a pointK(0 ≤K≤ 100,000) on the same number line. Farmer John has two mod...