标签：acm codeforces dp trie
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.
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.
If the player who moves first wins, print "First", otherwise print "Second" (without the quotes).
2 3 a b
First
3 1 a b c
First
1 2 ab
Second
给出N个字符，按照规则玩（太长了，这里就不翻译了）。然后第i-th输了的人，可以在第i+1-th先手，现在要求第k-th赢了的人是谁。
首先存这些字符，用trie来存，通过trie就很容易联想到树型DP，这里的DP就不是取最优值之类的了，而是用来弄到达某个节点的胜负情况。
我们首先设节点v，win[v]代表已经组装好的字符刚好匹配到v了，然后需要进行下一步匹配时，先手是否可以赢，lose[v]则代表先手是否会输。
叶节点，win[v] = false, lose[v] = true.
其他节点 win[v] = win[v] | !win[child], lose[v] = lose[v] | !lose[child]. (因为每次赢的人，下一个就不是先手，所以结果肯定是跟下一个节点的赢成对立关系)。
如若win[0] = true , lose[0] = true则意味着第一局的人可以操控结果，否则按照k的次数来判断是否可以赢。
#include <cstdio>
#include <cstring>
#define N_max 123456
#define sigma_size 26
using namespace std;
bool win[N_max], lose[N_max];
int n, k;
char st1[N_max];
class Trie{
public:
int ch[N_max][sigma_size];
int sz;
Trie() {
sz=0;
memset(ch[0], 0, sizeof(ch[0]));
}
int idx(char c) {
return c-'a';
}
void insert(char *s) {
int l = strlen(s), u = 0;
for (int i = 0; i < l; i++) {
int c = idx(s[i]);
if (!ch[u][c]) {
ch[u][c] = ++sz;
memset(ch[sz], 0, sizeof(ch[sz]));
}
u = ch[u][c];
}
}
};
Trie T;
void init() {
scanf("%d%d", &n, &k);
for (int i = 1; i <= n; i++) {
scanf("%s", st1);
T.insert(st1);
}
}
void dfs(int v) {
bool is_leaf = true;
win[v] = false;
lose[v] = false;
for (int i = 0; i < sigma_size; i++) {
int tmp = T.ch[v][i];
if (tmp) {
is_leaf = false;
dfs(T.ch[v][i]);
win[v] |= !win[T.ch[v][i]];
lose[v] |= !lose[T.ch[v][i]];
}
}
if (is_leaf) {
win[v] = false;
lose[v] = true;
}
}
void ans(bool res) {
puts(res? "First":"Second");
}
void solve() {
dfs(0);
if (win[0] && lose[0])
ans(true);
else if (win[0])
ans(k&1);
else
ans(0);
}
int main() {
init();
solve();
}codeforces Round #260(div2) D解题报告
标签：acm codeforces dp trie
原文地址：http://blog.csdn.net/lmhunter/article/details/40404563
