文章目录
题目描述
输入样例
15
chris smithm
adam smithm
bob adamsson
jack chrissson
bill chrissson
mike jacksson
steve billsson
tim mikesson
april mikesdottir
eric stevesson
tracy timsdottir
james ericsson
patrick jacksson
robin patricksson
will robinsson
6
tracy tim james eric
will robin tracy tim
april mike steve bill
bob adam eric steve
tracy tim tracy tim
x man april mikes
输出样例
Yes
No
No
Whatever
Whatever
NA
思路
AC代码
#include <bits/stdc++.h>
using namespace std;
typedef struct
{
char sex;
string family_name;
}person;
map<string, person> mp;
bool judge(string a, string b)
{
int cnta = 1; //统计A祖先的代数
for(string A = a; A.size(); A = mp[A].family_name)
{
int cntb = 1; //统计B祖先的代数
for(string B = b; B.size(); B = mp[B].family_name)
{
if(cnta >= 5 && cntb >= 5) return true; //五代之内没有祖先
if(A == B && (cnta < 5 || cntb < 5)) return false; //五代之内有相同祖先
cntb ++;
}
cnta ++;
}
return true; //不够五代
}
int main()
{
int n;
cin >> n;
for(int i = 0; i < n; i ++)
{
string s1, s2;
cin >> s1 >> s2;
if(s2.find("sson") != -1) //维京人后裔 男性
{
int pos = s2.find("sson");
mp[s1] = {'m', s2.substr(0, pos)};
}
else if(s2.find("sdottir") != -1) //维京人后裔 女性
{
int pos = s2.find("sdottir");
mp[s1] = {'f', s2.substr(0, pos)};
}
else
{
int len = s2.size();
if(s2[len - 1] == 'm') mp[s1].sex = 'm';
else if(s2[len - 1] == 'f') mp[s1].sex = 'f';
}
}
int m;
cin >> m;
while(m --)
{
string name1, family_name1, name2, family_name2;
cin >> name1 >> family_name1 >> name2 >> family_name2;
if(mp.find(name1) == mp.end() || mp.find(name2) == mp.end()) cout << "NA" << endl;
else if(mp[name1].sex == mp[name2].sex) cout << "Whatever" << endl;
else
{
if(judge(name1, name2)) cout << "Yes" << endl;
else cout << "No" << endl;
}
}
return 0;
}
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