codeforces 455B. A Lot of Games

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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 3ab
output
First
input
3 1abc
output
First
input
1 2ab
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|1bool 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_JUDGEfreopen("in.txt","r",stdin);//  freopen("out.txt","w",stdout);#endifint 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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