codeforces Round #260(div2) D解决报告
1 second
256 megabytes
standard input
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.
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
题目大意:
给出N个字符。依照规则玩(太长了。这里就不翻译了)。
然后第i-th输了的人,能够在第i+1-th先手,如今要求第k-th赢了的人是谁。
解法:
首先存这些字符。用trie来存,通过trie就非常easy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。然后须要进行下一步匹配时,先手能否够赢。lose[v]则代表先手是否会输。
叶节点,win[v] = false, lose[v] = true.
其它节点 win[v] = win[v] | !win[child], lose[v] = lose[v] | !lose[child]. (由于每次赢的人。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能够操控结果,否则依照k的次数来推断能否够赢。
代码:
#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();
}
版权声明:本文博主原创文章。博客,未经同意不得转载。
codeforces Round #260(div2) D解决报告的更多相关文章
- codeforces Round #259(div2) E解决报告
E. Little Pony and Summer Sun Celebration time limit per test 1 second memory limit per test 256 meg ...
- codeforces Round #259(div2) D解决报告
D. Little Pony and Harmony Chest time limit per test 4 seconds memory limit per test 256 megabytes i ...
- codeforces Round #258(div2) D解题报告
D. Count Good Substrings time limit per test 2 seconds memory limit per test 256 megabytes input sta ...
- codeforces Round #259(div2) C解题报告
C. Little Pony and Expected Maximum time limit per test 1 second memory limit per test 256 megabytes ...
- Codeforces Round #260(div2)C(递推)
有明显的递推关系: f[i]表示i为数列中最大值时所求结果.num[i]表示数i在数列中出现了几次. 对于数i,要么删i,要么删i-1,只有这两种情况,且子问题还是一样的思路.那么很显然递推一下就行了 ...
- codeforces Round #258(div2) C解题报告
C. Predict Outcome of the Game time limit per test 2 seconds memory limit per test 256 megabytes inp ...
- Codeforces Round#320 Div2 解题报告
Codeforces Round#320 Div2 先做个标题党,骗骗访问量,结束后再来写咯. codeforces 579A Raising Bacteria codeforces 579B Fin ...
- Codeforces Round #539 div2
Codeforces Round #539 div2 abstract I 离散化三连 sort(pos.begin(), pos.end()); pos.erase(unique(pos.begin ...
- 【前行】◇第3站◇ Codeforces Round #512 Div2
[第3站]Codeforces Round #512 Div2 第三题莫名卡半天……一堆细节没处理,改一个发现还有一个……然后就炸了,罚了一啪啦时间 Rating又掉了……但是没什么,比上一次好多了: ...
随机推荐
- windows server 搭建radius服务器
使用ISA 2004搭建PPTP/L2TP 服务器后,VPN账号是在radius服务器上的,使用window server 2003搭建radius服务器,需要添加组件->internet验证服 ...
- 2)JS动态生成HTML元素的爬取
2)JS动态生成HTML元素的爬取 import java.util.List; import org.openqa.selenium.By; import org.openqa.selenium.W ...
- PHP实现插入排序算法
插入排序(Insertion Sort),是一种较稳定.简单直观的排序算法.插入排序的工作原理,是通过构建有序序列,对于未排序的数据,在有序序列中从后向前扫描,找到合适的位置并将其插入.插入排序,在最 ...
- 在Java中如何使用jdbc连接Sql2008数据库(转)
我们在javaEE的开发中,肯定是要用到数据库的,那么在javaEE的开发中,是如何使用代码实现和SQL2008的连接的呢?在这一篇文章中,我将讲解如何最简单的使用jdbc进行SQL2008的数据库的 ...
- jquery 弹出登陆框,简单易懂!修改密码效果代码
在网上找了一大堆,看的眼花瞭乱,还是研究原码,自已搞出来了! ui原地址:http://jqueryui.com/dialog/#modal-form 可以把js,css下载到本地,要不然不联网的话, ...
- IT痴汉的工作现状13-吓唬电话
那是一个普通的周末上午,稍微阴沉的天,使得暑气消退了好多.刚吃过早饭,我懒懒的浏览着CSDN论坛上有趣的问题和答案. 突然电话响起.是一个陌生的号码.我像往常一样接起电话,""您好 ...
- HDU 1201
18岁生日 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Subm ...
- uva-442 Matrix Chain Multiplication
Suppose you have to evaluate an expression like A*B*C*D*E where A,B,C,D and E are matrices. Since ma ...
- 性能是全新的 SEO
作为一个前端project师,那不只就是公开地处理那些美丽的html5, css3 和javascript特效.小而重要的一部分工作就是要让项目朝着代码稳定和代码标准方向进展.设计.信息结构以及后台限 ...
- 栈实现java
栈是一种“先去后出”的抽象的数据结构.例如:我们在洗盘子的时候,洗完一个盘子,将其放在一摞盘子的最上面,但我们全部洗完后,要是有盘子时,我们会先从最上面的盘子开始使用,这种例子就像栈的数据结构一样,先 ...