天梯赛 L2 愿天下有情人都是失散多年的兄妹

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/ccDLlyy/article/details/79808222

题目链接：点击打开链接

思路：反向建树，搜索两个人各自5层以内的祖先，看有没有交集即可。坑点是，一定要保存父母辈的信息，题目的测试用例有判断父母辈的能不能成为情侣......

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm> 
#include <set>
#include <queue>
using namespace std;
#define M 100000
int le[M],rig[M],sex[M];
set<int> ance[3];
void bfs(int x,int loc){
	queue<int> q;
	q.push(x);
	int level = 1,last = x;
	while(!q.empty()){
		int t = q.front();
		q.pop();
		ance[loc].insert(t);
		if(le[t] != -1) q.push(le[t]);
		if(rig[t] != -1) q.push(rig[t]);
		if(t == last){
			level++;
			last = q.back();
			if(level == 6) break;
		}
	}
	return;
}
int main(){
	int n;
	int id,fa,ma;
	char se;
	memset(le,-1,sizeof(le));
	memset(rig,-1,sizeof(rig));
	scanf("%d",&n);
	for(int i = 0;i < n;i++){
		scanf("%d",&id);
		getchar();
		se = getchar();
		scanf("%d%d",&fa,&ma);
		sex[id] = (se == 'M'?1:0);
		le[id] = fa;
		rig[id] = ma;
		if(fa != -1){//记录父母性别 
			sex[fa] = 1;
		}
		if(ma != -1){
			sex[ma] = 0;
		}
	}
	int q,x,y;
	scanf("%d",&q);
	while(q--){
		scanf("%d%d",&x,&y);
		if(sex[x] == sex[y]){
			printf("Never Mind\n");
			continue;
		}
		ance[0].clear();
		ance[1].clear();
		ance[2].clear();
		bfs(x,0);
		bfs(y,1);
		set_intersection(ance[0].begin(),ance[0].end(),ance[1].begin(),ance[1].end(),inserter(ance[2],ance[2].begin()));
		puts((int)ance[2].size()?"No":"Yes");
	} 
	return 0;
}