Codeforces 456 Problem D A Lot of Games

最新推荐文章于 2021-10-31 22:14:35 发布

xasd2003

最新推荐文章于 2021-10-31 22:14:35 发布

阅读量500

点赞数

CC 4.0 BY-SA版权

文章标签： codeforces

本文链接：https://blog.youkuaiyun.com/u013659964/article/details/38580531

本文探讨了游戏开发中涉及的策略与算法实现，包括玩家交互、AI行为、游戏优化等核心内容。通过深入分析游戏设计中的关键元素，提供了一套优化游戏体验的技术解决方案。

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).

题意是说给定n个字符串，两个人做k次游戏，每一次游戏的规则是，首先有一个空串，两个人轮流向串中添加字符，必须满足这个字符串是给定的n个字符串中的其中一个的前缀，如果有人不能添加字符，则说明他输了这一次，下一次游戏由输的人先手，直到k次游戏结束，如果先手赢了输出first，否则输出second；

对于先手来说，

如果先手不仅有必胜的策略，而且有必败的策略，他可以先输k-1盘，然后赢一盘就可以了，先手赢；

如果先手有必胜的策略，没有必败的策略，那么两个人轮流先手，如果有奇数盘，则先手赢，偶数盘，则后手赢；

如果只有必败的策略，那么后手可以一直让他败下去，则后手赢；

我们可以利用n个字符串建立一个Trie树，用两个数组记录先手是否有必胜态，或者必败态，拿必胜态的数组举例子，因为先手只能拿到奇数层，后手拿到偶数层，如果这个点是偶数层的，那么它的子节点是先手取，如果子节点中存在必胜态，则这个节点也是必胜态。如果这个点是奇数层，子节点中存在必败态，则这个节点也是必败的。这样一直更新到根节点，得到先手是否有必胜的策略然后再判断是否有必败的策略就可以了。

#include<cstdio>
#include<cmath>
#include<cstring>
const int Maxn=100009;
int chd[Maxn][26],pre[Maxn],State[Maxn],State2[Maxn],level[Maxn];
int p,nn=1;
char str[Maxn];

void insert(char *str)
{
	p=0;
	int cnt=0;
	while(*str)
	{
		cnt++;
		int c=(*str++)-'a';
		if(!chd[p][c])
		{
			memset(chd[nn],0,sizeof(chd[nn]));
			chd[p][c]=nn++;
			
		}
		pre[chd[p][c]]=p;
		p=chd[p][c];
		level[p]=(cnt&1);
	}
	State[p]=State2[p]=(cnt&1)+1;
}

int main()
{
	
	int n,k;
	scanf("%d%d",&n,&k);
	memset(chd[0],0,sizeof(chd[0]));
	for(int i=0;i<n;i++)
	{
		scanf("%s",str);
		insert(str);
	}
	
	
	for(int i=nn-1;i>0;i--)
	{
		int u=pre[i];
		if(!State[u]) State[u]=State[i];
		else
		{
			if(level[u]&&State[i]==1) State[u]=1;
			if(!level[u]&&State[i]==2) State[u]=2;
		}
	}
	for(int i=nn-1;i>0;i--)
	{
		int u=pre[i];
		if(!State2[u]) State2[u]=State2[i];
		else
		{
			if(!level[u]&&State2[i]==1) State2[u]=1;
			if(level[u]&&State2[i]==2) State2[u]=2;
		}
	}
	        
    
	if(State[0]==2)
	{
		if(State2[0]==1) printf("First\n");
		else 	
		{
			printf("%s\n", (k & 1) ? "First" : "Second");
		}
	}
	else printf("Second\n");
	
	return 0;
}