PTA 7-27 家谱处理 (30 point(s))

#include<bits/stdc++.h>
using namespace std;

string pre[100];               // 缓存相对关系
map<string, string> parent;     // 存储对象父亲

bool isAncestor(string x, string y){
    // x不是y的祖先 && y未到祖先(root祖先父节点)
    while(x != parent[y] && parent[y] != "root")
        y = parent[y];
    // 返回判断结果
    return x == parent[y];
}

bool judge(string x, string y, char r){
    if(r == 'c') return parent[x] == y;
    else if(r == 'p') return x == parent[y];
    else if(r == 's') return parent[x] == parent[y];
    else if(r == 'a') return isAncestor(x, y);
    else if(r == 'd') return isAncestor(y, x);
}

int main() {
    int n, m;
    cin >> n >> m;
    getchar();
    while(n--){
        string name;
        getline(cin, name);
        int space = count(name.begin(), name.end(), ' ');
        // 如果空格数相同则覆盖
        name = name.substr(space);
        pre[space / 2] = name;
        // 空格为0是祖先 否则读取数组上一个存储的元素
        parent[name] = (space == 0 ? "root" : pre[space / 2 - 1]);
    }
    while(m--){
        string x, y, omit, relation;
        cin >> x >> omit >> omit >> relation >> omit >> y;
        cout << (judge(x, y, relation[0]) == true ? "True" : "False") << endl;
    }
}

Robert is an ancestor of Andrew 

示例这一行，Robert和Andrew至隔了一代关系，判断结果算对。所以一层关系也算祖先关系。 

参考代码        参考理解