Fennec VS. Snuke --AtCoder
题目描述
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.
输入
N
a1 b1
:
aN−1 bN−1
输出
样例输入
7
3 6
1 2
3 1
7 4
5 7
1 4
样例输出
Fennec
分析:本题最优策略是尽量堵住对手,BFS扫到最后发现谁占据的节点多,谁就会赢。
#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#define range(i,a,b) for(int i=a;i<=b;++i)
#define LL long long
#define rerange(i,a,b) for(int i=a;i>=b;--i)
#define fill(arr,tmp) memset(arr,tmp,sizeof(arr))
using namespace std;
int n,x,y,col[],cnt[];
vector<int> map[];
void init(){
cin>>n;
range(i,,n-){
cin>>x>>y;
map[x].push_back(y);
map[y].push_back(x);
}
col[]=,col[n]=;
}
void solve(){
queue<int>q;
q.push(),q.push(n);
while(!q.empty()){
int head=q.front();
q.pop();
++cnt[col[head]];
range(i,,map[head].size()-){
int tmp=map[head][i];
if(col[tmp])continue;
col[tmp]=col[head];
q.push(tmp);
}
}
cout<<(cnt[]<cnt[]?"Fennec":"Snuke")<<endl;
}
int main() {
init();
solve();
return ;
}
Fennec VS. Snuke --AtCoder的更多相关文章
- AtCoder Beginner Contest 067 D - Fennec VS. Snuke
D - Fennec VS. Snuke Time limit : 2sec / Memory limit : 256MB Score : 400 points Problem Statement F ...
- 【AtCoder078D】Fennec VS. Snuke
AtCoder Regular Contest 078 D - Fennec VS. Snuke 题意 给一个树,1是白色,n是黑色,其它没有颜色.Fennec每次可以染白色点的直接邻居为白色.Snu ...
- Fennec VS. Snuke
Fennec VS. Snuke Time limit : 2sec / Memory limit : 256MB Score : 400 points Problem Statement Fenne ...
- ABC Fennec VS. Snuke
题目描述 Fennec and Snuke are playing a board game. On the board, there are N cells numbered 1 through N ...
- ARC078 D.Fennec VS. Snuke(树上博弈)
题目大意: 给定一棵n个结点的树 一开始黑方占据1号结点,白方占据n号结点 其他结点都没有颜色 每次黑方可以选择黑色结点临近的未染色结点,染成黑色 白方同理. 最后谁不能走谁输. 题解: 其实简单想想 ...
- AtCoder Regular Contest 078
我好菜啊,ARC注定出不了F系列.要是出了说不定就橙了. C - Splitting Pile 题意:把序列分成左右两部分,使得两边和之差最小. #include<cstdio> #inc ...
- 【AtCoder】ARC078
C - Splitting Pile 枚举从哪里开始分的即可 #include <bits/stdc++.h> #define fi first #define se second #de ...
- AtCoder Regular Contest 078 D
D - Fennec VS. Snuke Time limit : 2sec / Memory limit : 256MB Score : 400 points Problem Statement F ...
- すぬけ君の塗り絵 / Snuke's Coloring AtCoder - 2068 (思维,排序,贡献)
Problem Statement We have a grid with H rows and W columns. At first, all cells were painted white. ...
随机推荐
- 4 Template层- HTML转义
1.HTML转义 Django对字符串进行自动HTML转义,如在模板中输出如下值: 视图代码: def index(request): return render(request, 'temtest/ ...
- Python框架之Django学习笔记(九)
模型 之前,我们用 Django 建造网站的基本途径: 建立视图和 URLConf . 正如我们所阐述的,视图负责处理一些主观逻辑,然后返回响应结果. 作为例子之一,我们的主观逻辑是要计算当前的日期和 ...
- postgresql connection failure:SQLSTATE[08006] [7] could not connect to server: Permission denied Is the server running on host "127.0.0.1" and accepting TCP/IP connections on port 5432?
PHP 程序无法连接到 CentOS 上的PostgreSQL,但是在 CentOS 服务器上却能正常运行 psql, 操作如下:多次重启 PG 数据库后发现 CGI 脚本无法连接数据库,但是可以使用 ...
- [oldboy-django][2深入django]django模板使用函数
1 模板引入子html--include 模板引擎 - 母版 - include,导入公共的html a. 用法:{% include "pub.html" %}, pub.htm ...
- TOJ 3750: 二分查找
3750: 二分查找 Time Limit(Common/Java):3000MS/9000MS Memory Limit:65536KByteTotal Submit: 1925 ...
- mysql数据库二进制初始化出现:170425 17:47:04 [ERROR] /application/mysql//bin/mysqld: unknown option '--skip-locking' 170425 17:47:04 [ERROR] Aborting 解决办法
[root@localhost mysql]# ./scripts/mysql_install_db --user=mysql --basedir=/application/mysql/ --data ...
- BZOJ 1855: [Scoi2010]股票交易(DP+单调队列)
1855: [Scoi2010]股票交易 Description 最近lxhgww又迷上了投资股票,通过一段时间的观察和学习,他总结出了股票行情的一些规律. 通过一段时间的观察,lxhgww预测到了未 ...
- POJ 3249:Test for Job(拓扑排序+DP)
题意就是给一个有向无环图,每个点都有一个权值,求从入度为0的点到出度为0点路径上经过点(包括起点终点)的权值和的最大值. 分析: 注意3点 1.本题有多组数据 2.可能有点的权值是负数,也就是结果可能 ...
- 【bzoj2048】[2009国家集训队]书堆 数论
题目描述 输入 第一行正整数 N M 输出 一行(有换行符),L,表示水平延伸最远的整数距离 (不大于答案的最大整数) 样例 #1 Input: 1 100 Output: 49 #2 Input: ...
- BZOJ 4561 [JLoi2016]圆的异或并 ——扫描线
扫描线的应用. 扫描线就是用数据结构维护一个相对的顺序不变,带修改的东西. 通常只用于一次询问的情况. 抽象的看做一条垂直于x轴直线从左向右扫过去. 这道题目要求求出所有圆的异或并. 所以我们可以求出 ...