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 ;
}
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