AtCoder Beginner Contest 067 D - Fennec VS. Snuke
D - Fennec VS. Snuke
Time limit : 2sec / Memory limit : 256MB
Score : 400 points
Problem Statement
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N−1 roads, each connecting two cells. Cell ai is adjacent to Cell bi through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
- Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
- Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
- 2≤N≤105
- 1≤ai,bi≤N
- The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a1 b1
:
aN−1 bN−1
Output
If Fennec wins, print Fennec; if Snuke wins, print Snuke.
Sample Input 1
7
3 6
1 2
3 1
7 4
5 7
1 4
Sample Output 1
Fennec
For example, if Fennec first paints Cell 2 black, she will win regardless of Snuke's moves.
Sample Input 2
4
1 4
4 2
2 3
Sample Output 2
Snuke
F先走,S后走,且只能走相邻的点,所以,bfs搜索所有的点,s可以走的点不少于f是就获胜
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define PI 3.141592653589793238462
#define INF 1000000000
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
vector<int>v[];
int x,y,n;
int ans[]={,,};
int vis[];
void bfs(int x,int y)
{
mem(vis);
vis[x]=;
vis[y]=;
queue<int>q;
q.push(x);
q.push(y);
while(!q.empty())
{
int pos=q.front();
ans[vis[pos]]++;
q.pop();
for(int i=;i<v[pos].size();i++)
{
if(vis[v[pos][i]]) continue;
vis[v[pos][i]]=vis[pos];
q.push(v[pos][i]);
}
}
}
int main()
{
scanf("%d",&n);
for(int i=;i<n;i++)
{
scanf("%d%d",&x,&y);
v[y].push_back(x);
v[x].push_back(y);
}
bfs(,n);
puts(ans[]>=ans[]?"Snuke":"Fennec");
}
AtCoder Beginner Contest 067 D - Fennec VS. Snuke的更多相关文章
- AtCoder Beginner Contest 067 C - Splitting Pi
C - Splitting Pile Time limit : 2sec / Memory limit : 256MB Score : 300 points Problem Statement Snu ...
- AtCoder Beginner Contest 100 2018/06/16
A - Happy Birthday! Time limit : 2sec / Memory limit : 1000MB Score: 100 points Problem Statement E8 ...
- AtCoder Beginner Contest 053 ABCD题
A - ABC/ARC Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Smeke has ...
- AtCoder Beginner Contest 068 ABCD题
A - ABCxxx Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement This contes ...
- AtCoder Beginner Contest 154 题解
人生第一场 AtCoder,纪念一下 话说年后的 AtCoder 比赛怎么这么少啊(大雾 AtCoder Beginner Contest 154 题解 A - Remaining Balls We ...
- AtCoder Beginner Contest 153 题解
目录 AtCoder Beginner Contest 153 题解 A - Serval vs Monster 题意 做法 程序 B - Common Raccoon vs Monster 题意 做 ...
- AtCoder Beginner Contest 052
没看到Beginner,然后就做啊做,发现A,B太简单了...然后想想做完算了..没想到C卡了一下,然后还是做出来了.D的话瞎想了一下,然后感觉也没问题.假装all kill.2333 AtCoder ...
- AtCoder Beginner Contest 136
AtCoder Beginner Contest 136 题目链接 A - +-x 直接取\(max\)即可. Code #include <bits/stdc++.h> using na ...
- AtCoder Beginner Contest 137 F
AtCoder Beginner Contest 137 F 数论鬼题(虽然不算特别数论) 希望你在浏览这篇题解前已经知道了费马小定理 利用用费马小定理构造函数\(g(x)=(x-i)^{P-1}\) ...
随机推荐
- docker环境下mysql参数修改
原文:docker环境下mysql参数修改 需要修改log_bin为on,看了好几个博客说都需要删掉容器重新生成,然而并非如此, 我们可以用docker cp 命令将docker的文件"下载 ...
- Codeforces Round #256 (Div. 2) B. Suffix Structures(模拟)
题目链接:http://codeforces.com/contest/448/problem/B --------------------------------------------------- ...
- python之路-------字符串与正則表達式
1.1.#####去掉字符串中的转义符string.strip() print "hello\tworld\n" >>> word="\thello w ...
- Vue源代码笔记(一)数据绑定
VUE数据绑定介绍 数据绑定是vue的基础核心之一,本文以Vue对象(当然也包含VueComponent)里的data数据绑定为例,阐述整个绑定的过程. Vue的数据绑定由三部分组成, Observe ...
- TYVJ 1340 折半暴搜+二分
思路: 1. 这 题 不卡常过不去啊-- (先加一个random_shuffle) 首先 我们可以折半 搜这一半可以到达的重量 sort一遍 然后搜另一半 对于路程中每一个解 我们可以二分前一半中加这 ...
- UVa 1151 Buy or Build【最小生成树】
题意:给出n个点的坐标,现在需要让这n个点连通,可以直接在点与点之间连边,花费为两点之间欧几里得距离的平方,也可以选购套餐,套餐中所含的点是相互连通的 问最少的花费 首先想kruskal算法中,被加入 ...
- sql sever 创建临时表的两种方法
创建临时表 方法一: create table #临时表名( 字段1 约束条件, 字段2 约束条件, .....) ...
- session 存入 redis
<?php header('content-type:text/html;charset=utf-8'); /* * 更改 session 存储位置及存储方式. */ ini_set('sess ...
- 学习Go语言之观察者模式
首先了解一下观察者模式 1.目标和观察者抽象对象需要首先建立 //抽象主题 type Subject interface { Add(o Observer) Send(str string) } // ...
- PostgreSQL指定用户可访问的数据库pg_hba.conf
进入指定目录: # cd /var/lib/pgsql/9.3/data/ 使用vi编辑pg_hba.conf文件 # vi pg_hba.conf 以上配置为所有IP及网关都允许访问,使用MD5认证 ...