AtCoder：Fennec VS. Snuke（dfs & 思維）

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

Copy

7
3 6
1 2
3 1
7 4
5 7
1 4
      

Sample Output 1

Copy

Fennec
      

For example, if Fennec first paints Cell 2 black, she will win regardless of Snuke's moves.

Sample Input 2

Copy

4
1 4
4 2
2 3
      

Sample Output 2

Copy

Snuke      

題意：一棵樹由N個節點，1節點黑色，N節點白色，兩個人輪流塗色，先手将黑色節點相鄰一個無色節點塗成黑色，後手白色亦然，誰不能塗就輸，問誰會赢。

思路：最優政策顯然先往對方的路徑塗過去，那麼計算一下兩色相遇時誰的子節點多即可。

# include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+3;
int cnt=0, id=0, Next[maxn],in[maxn],out[maxn];
int fa[maxn][20], len[maxn];
struct node
{
    int v, next;
}edge[maxn<<1];
void add_edge(int u, int v)
{
    edge[cnt] = {v, Next[u]};
    Next[u] = cnt++;
    edge[cnt] = {u, Next[v]};
    Next[v] = cnt++;
}
 
void dfs(int cur, int pre, int d)
{
    fa[cur][0] = pre;
    len[cur] = d;
    for(int i=1; i<20; ++i)
        fa[cur][i] = fa[fa[cur][i-1]][i-1];
    in[cur] = ++id;
    for(int i=Next[cur]; i!=-1; i=edge[i].next)
    {
        int v = edge[i].v;
        if(v == pre) continue;
        dfs(v, cur, d+1);
    }
    out[cur] = id;
}
int main()
{
    int n, a, b;
    scanf("%d",&n);
    memset(Next, -1, sizeof(Next));
    for(int i=1; i<n; ++i)
    {
        scanf("%d%d",&a,&b);
        add_edge(a, b);
    }
    dfs(1, 0, 0);
    int tmp=(len[n]-1)/2, now=n;
    for(int i=0; i<20; ++i)
        if(tmp>>i&1) now = fa[now][i];
    int bl = n-(out[now]-in[now]+1)-len[n]/2-1;
    int wh = out[now]-in[now]+1-tmp-1;
    if(len[n]&1)
    {
        if(bl <= wh) puts("Snuke");
        else puts("Fennec");
    }
    else
    {
        if(wh <= bl) puts("Fennec");
        else puts("Snuke");
    }
    return 0;
}