http://poj.org/problem?id=3278 从n出发,向两边转移,为了不使数字无限制扩大,限制在2*k以内, 注意不能限制在k以内,否则就缺少不断使用-1得到的一些结果 #include <cstdio> #include <cstring> #include <algorithm> #include <queue> using namespace std; const int maxn=2e5+3; int n,k; int dp[max…

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 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…

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: 2000MS   Memory Limit: 65536K Total Submissions: 61826   Accepted: 19329 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 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…

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…

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…

#include<stdio.h> #include<string.h> #include<algorithm> #include<queue> using namespace std; ],vis[]; queue<int>q; int bfs(int n,int k) { int head,next,i; q.push(n); a[n]=; vis[n]=; ) { head=q.front(); q.pop(); ;i<;i++) {…

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…