冰岛人[天梯赛]

文章目录

题目描述
思路
AC代码

题目描述

在这里插入图片描述

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

模拟

存储结构
1.结构体存储每个人的信息，包括父亲的姓名（只针对维京人后裔）以及每个人的性别
2.用map存储每个人姓名到其结构体信息的映射

具体流程
1.判断两人是否都存在 – 一定要首先判断
2.判断两人性别是否相同
3.判断两人是否五代内有公共祖先
①依次遍历每个人的祖先，判断五代内是否有公共祖先，如果超出五代，或者祖先不足5代（不管相同还是不用），则可以交往；否则不可以交往

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;
}