Codeforces Gym 100814J Game 简单博弈

题目大意：

就是现在初始的时候有一个字符串S, 长度不超过10000

两个人轮流操作

每次可以选择从左向右将连续两个字符按照给出的矩阵进行变化

或者选择从右往左

如果最后剩下的是元音字母则Salah获胜, 否则Marzo获胜

Salah先手问谁会获胜

大致思路：

比较水, 考虑到S在每次变换之后长度都会接近减半, 于是可以直接dfs暴力搜索

记下当前串以及当前轮到谁即可

代码如下：

Result  :  Accepted     Memory  :  148 KB     Time  :  31 ms

/*
* Author: Gatevin
* Created Time: 2015/11/21 16:25:41
* File Name: Sakura_Chiyo.cpp
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;

int T;
char s[30][30];

char get(char c1, char c2)
{
return s[c1 - 'a'][c2 - 'a'];
}

bool isVowel(char c)
{
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}

int dfs(string now, char who)
{
int sz = now.size();
if(sz == 1 && isVowel(now[0]))
return 1;
if(sz == 1 && !isVowel(now[0]))
return 0;
string left , right;
for(int i = 0; i + 1 < sz; i += 2)
left.push_back(get(now[i], now[i + 1]));
if(sz & 1) left.push_back(now[sz - 1]), right.push_back(now[0]);
for(int i = (sz & 1); i < sz; i += 2)
right.push_back(get(now[i], now[i + 1]));
int L = dfs(left, who == 's' ? 'm' : 's'), R = dfs(right, who == 's' ? 'm' : 's');
if(who == 's')
{
if(L == 1 || R == 1) return 1;
return 0;
}
else
{
if(L == 0 || R == 0) return 0;
return 1;
}
}

int main()
{
ios::sync_with_stdio(0);
cin>>T;
while(T--)
{
string tmp;
for(int i = 0; i < 26; i++)
{
cin>>tmp;
for(int j = 0; j < 26; j++)
s[i][j] = tmp[j];
}
string in;
cin>>in;
printf("%s\n", dfs(in, 's') ? "Salah" : "Marzo");
}
return 0;
}
/*

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*/