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Question 246. 2022 Winter Ladder Training - 7-5 Icelanders (25 points)
1. The topic
2. Problem solving
When I did this question before, I always thought without thinking that the requirement other than five generations was that as long as the elders within five generations of one party did not have the elders of the other party, it would be fine, and then just die directly, which is only correct test point 0. . . Only later did I discover that the original requirement was that if there was such a common ancestor, he had to satisfy that both parties were five generations and beyond. That is to say, in the case of No, the common ancestor is within five generations of both parties or within five generations of one party.
//五代以外包括五代必须是要满足任何一方的
#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+1;
map<string,int> index0;//以人的名为key，编号为value
int parent[maxn];//表示下标对应的人的父辈
int flag[maxn];//用于存储后序寻祖对应的祖先的代数
struct People
{
string name;
string familyname;
int sex;
}P[maxn];
int main()
{
fill(parent,parent+maxn,-1);
int N;
cin>>N;
for(int i=0;i<N;i++)
{
cin>>P[i].name>>P[i].familyname;
index0[P[i].name]=i;
}
for(int i=0;i<N;i++)
{
int pos_m,pos_f;
string fa;
pos_m=P[i].familyname.find("sson");
if(pos_m!=-1)
{
fa=P[i].familyname.substr(0,pos_m);
P[i].sex=1;
}
pos_f=P[i].familyname.find("sdottir");
if(pos_f!=-1)
{
fa=P[i].familyname.substr(0,pos_f);
P[i].sex=0;
}
if(pos_m==-1&&pos_f==-1)
{
if(P[i].familyname[P[i].familyname.length()-1]=='m')
{
P[i].sex=1;
}
else
{
P[i].sex=0;
}
}
else
{
parent[i]=index0[fa];
}
}
int M;
cin>>M;
for(int i=0;i<M;i++)
{
string name1,familyname1;
string name2,familyname2;
cin>>name1>>familyname1>>name2>>familyname2;
int index1,index2;
if(index0.find(name1)==index0.end()||index0.find(name2)==index0.end())//查无此人
{
printf("NA\n");
continue;
}
else
{
index1=index0[name1];
index2=index0[name2];
}
if(P[index1].sex==P[index2].sex)//同性
{
printf("Whatever\n");
continue;
}
fill(flag,flag+maxn,0);
int cnt;
cnt=1;
while(parent[index1]!=-1)//寻index1这个人的祖并存代
{
flag[index1]=cnt;//index1记为1代，这个必须标记，因为有可能这个index1就是index2这个人的祖先
index1=parent[index1];
cnt++;
}
flag[index1]=cnt;
cnt=1;
int ans=1;
while(parent[index2]!=-1)//寻index2这个人的祖并比对两个人祖宗的代数
{
//No情况包括双方都有五代以内或者一方有五代以内
if((flag[index2]>0&&flag[index2]<5&&cnt<5)||(flag[index2]>0&&flag[index2]<5&&cnt>=5)||(flag[index2]>=5&&cnt<5))
{
ans=0;
break;
}
index2=parent[index2];
cnt++;
}
if((flag[index2]>0&&flag[index2]<5&&cnt<5)||(flag[index2]>0&&flag[index2]<5&&cnt>=5)||(flag[index2]>=5&&cnt<5))//针对index2这个人的老祖设的判断
{
ans=0;
}
if(ans)
{
printf("Yes\n");
}
else
{
printf("No\n");
}
}
}