AtCoder-RC078 D - Fennec VS. Snuke（博弈）

最新推荐文章于 2019-06-28 08:41:00 发布

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本文介绍了一个关于Fennec与Snuke玩棋盘游戏的问题，玩家轮流涂色未被染色的格子，目标是使己方颜色占据更多位置。文章分析了最优策略并给出了实现代码。

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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.

此题比较简单，因为是树，那么只需要考虑从1号点到n号点的路径上的点的数量，最优策略是优先取这条路径上的点，所以将1号点当成根遍历，从n回溯到1的过程中，找出哪些点是Snuke可以涂色的，将最后一个Snuke可以得到的点当根，最后一个Fennec可以得到的点所在子树的大小就是Fennec可以涂色的所有点，比较即可。

附上代码

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#define N 654321
using namespace std;
int n,rt,size[N],fa[N],cnt=0,sum1,sum2;
int TOT,EN[N],LA[N],NE[N];
void ADD(int x,int y)
{
	TOT++;
	EN[TOT]=y;
	NE[TOT]=LA[x];
	LA[x]=TOT;
}
void DFS(int u,int f)
{
	int i,v;
	fa[u]=f;
	size[u]=1;
	for(i=LA[u];i;i=NE[i])
	{
		v=EN[i];
		if(v!=f)
		{
			DFS(v,u);
			size[u]+=size[v];
		}
	}
}
int main()
{
	int i,x,y;
	scanf("%d",&n);
	for(i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		ADD(x,y);ADD(y,x);
	}
	DFS(1,0);i=n;
	while(i!=1)i=fa[i],cnt++;
	cnt--;i=0;x=n;
	while(i<cnt/2)x=fa[x],i++;
	y=fa[x];
	DFS(x,0);
	sum1=size[y];sum2=n-sum1;
	if(sum1>sum2)cout<<"Fennec";
	else cout<<"Snuke";
}