codeforces 455B. A Lot of Games

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。

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B. A Lot of Games
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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 ≤ 1051 ≤ 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;
}




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