codeforces 455B. A Lot of Games

最新推荐文章于 2022-02-25 23:41:44 发布

原创最新推荐文章于 2022-02-25 23:41:44 发布 · 730 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#树

树字符串专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一个关于字符串的游戏，两个玩家轮流添加字符形成单词，探讨了最优策略下赢家预测的方法，并通过树形结构实现算法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Andrew, Fedor and Alex are inventive guys. Now they invent the game with strings for two players.

Given a group of n non-empty strings. During the game two players build the word together, initially the word is empty. The players move in turns. On his step player must add a single letter in the end of the word, the resulting word must be prefix of at least one string from the group. A player loses if he cannot move.

Andrew and Alex decided to play this game k times. The player who is the loser of the i-th game makes the first move in the (i + 1)-th game. Guys decided that the winner of all games is the player who wins the last (k-th) game. Andrew and Alex already started the game. Fedor wants to know who wins the game if both players will play optimally. Help him.

Input

The first line contains two integers, n and k (1 ≤ n ≤ 10⁵; 1 ≤ k ≤ 10⁹).

Each of the next n lines contains a single non-empty string from the given group. The total length of all strings from the group doesn't exceed 10⁵. Each string of the group consists only of lowercase English letters.

Output

If the player who moves first wins, print "First", otherwise print "Second" (without the quotes).

Sample test(s)

input
2 3
a
b

output
First

input
3 1
a
b
c

output
First

input
1 2
ab

output
Second

题意：两个人玩游戏，每次添一个字母，不能添的为输。输的在下一局中先手。问第k局谁赢。

挺简单的一道树形题，从后往前递推，判断初始点能不能输，能不能赢。

若不能赢，则先手一直输下去。

若可以赢，则看能不能输：

若能输，则先手赢（前k-1局一直输，第k局反杀）；

若不能输，则看局数，奇数局先手赢，偶数局后手赢。

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <functional>
#include <sstream>
#include <iomanip>
#include <cmath>
#include <cstdlib>
#include <ctime>
typedef long long ll;
//#pragma comment(linker, "/STACK:1024000000,1024000000")  //手动扩栈
#define INF 1e9
#define maxn 110000
#define maxm 100086+10
#define mod 1000000007
#define eps 1e-7
#define PI acos(-1.0)
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define scan(n) scanf("%d",&n)
#define scanll(n) scanf("%I64d",&n)
#define scan2(n,m) scanf("%d%d",&n,&m)
#define scans(s) scanf("%s",s);
#define ini(a) memset(a,0,sizeof(a))
#define out(n) printf("%d\n",n)
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
bool canWin[maxn],canLose[maxn];
struct Trie
{	
	int ch[maxn][26];
	int sz;
	Trie(){ init(); }
	void init()
	{
		sz = 1; memset(ch[0],0,sizeof(ch[0]));
	}
	int idx(char c) { return c - 'a'; }
	void insert(char *s)
	{
		int u = 0, n = strlen(s);
		rep(i,n)
		{
			int c = idx(s[i]);
			if(!ch[u][c])
			{
				memset(ch[sz],0,sizeof(ch[sz]));
				ch[u][c] = sz++;
			}
			u = ch[u][c];
		}
	}
	void solve()
	{
		for(int i = sz-1; i >= 0; i--)
		{
			canWin[i] = false;
			canLose[i] = count(ch[i],ch[i] + 26, 0) == 26;
			rep(j,26)
			{
				if(ch[i][j] == 0) continue;
				canWin[i] |= !canWin[ch[i][j]]; //canWin表示处在该点走的可以赢
				canLose[i] |= !canLose[ch[i][j]]; //canLose表示处在该点走的可以输
			}
		}
	}
}trie;
char s[maxn];
int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt","r",stdin);
	//  freopen("out.txt","w",stdout);
#endif
	int n,k;
	while(~scanf("%d%d",&n,&k))
	{
		trie.init();
		rep1(i,n)
		{
			scans(s);
			trie.insert(s);
		}
		trie.solve();
		
		if(!canWin[0]) { puts("Second"); continue; } //如果先手不能赢，则他将一直输下去
		if(canLose[0]) { puts("First"); continue; } //如果先手可以赢，也可以输，则他故意一直输，直到最后一局反杀

		if(k & 1) puts("First"); //如果先手只能赢，则看玩的次数，奇数则先手赢，偶数则后手赢
		else puts("Second");
	}
	return 0;
}