在并查集的基础上进行多支路的修改
#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <vector>
using namespace std;
//本题父亲母亲都有,也没有说是独生子女,怎么说 强行递归?
//可以用一个visit数组来进行剪枝 每个人都是大写英文字母
const int MAXN = 30;
bool visit[MAXN];
class parent{
public:
int mother;
int father;
parent():mother(-1),father(-1){}
parent(int a, int b):mother(a),father(b){}
};
int Find(int x, vector<parent> &v, int target, int cengshu){//初步设计:没遇到就返回0,遇到了就返回寻找层数
if(visit[x]){
return 0;
}else{
visit[x] = true;
}
if(x == target){
return cengshu;
}
if(v[x].mother == -1 && v[x].father == -1){
return 0;
}
if(v[x].mother != -1){
int direction1 = Find(v[x].mother, v, target, cengshu+1);
if(direction1){//说明往母亲方向找到了
return direction1;
}
}
if(v[x].father != -1){
return Find(v[x].father, v, target, cengshu+1);//母亲那边没过的话,这边过了就是过了,没过就没过
}
return 0;
}
void Union(int x, int y, int z, vector<parent> &v){
if(y != -1){
v[x].mother = y;
}
if(z != -1){
v[x].father = z;
}
}
int main(){
int n, m;
while(scanf("%d%d", &n, &m) != EOF){
memset(visit, false, sizeof(visit));
vector<parent> v;
for(int i=0; i<26; i++){
v.push_back(parent());
}
string s;
for(int i=0; i<n; i++){
cin >> s;
int y, z;
if(s[1] != '-'){
y = s[1]-'A';
}else{
y = -1;
}
if(s[2] != '-'){
z = s[2]-'A';
}else{
z = -1;
}
Union(s[0]-'A', y, z, v);
}
for(int i=0; i<m; i++){//比如输入FA,是问F是A的谁
cin >> s;
int find1 = Find(s[0]-'A', v, s[1]-'A', 0);
memset(visit, false, sizeof(visit));
if(find1){//从F往上找找到了A,说明F是A的后辈
while(find1 > 2){
printf("great-");
find1--;
}
if(find1 == 2){
printf("grandchild\n");
}else{
printf("child\n");
}
continue;
}
int find2 = Find(s[1]-'A', v, s[0]-'A', 0);
memset(visit, false, sizeof(visit));
if(find2){//从A往上找找到了F,说明F是A的祖辈
while(find2 > 2){
printf("great-");
find2--;
}
if(find2 == 2){
printf("grandparent\n");
}else{
printf("parent\n");
}
continue;
}
printf("-\n");
}
}
return 0;
}