hdu 4409 Family Name List

最新推荐文章于 2022-02-25 19:38:17 发布

原创最新推荐文章于 2022-02-25 19:38:17 发布 · 804 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#sorting #iterator #ini #string #c #output

Simulation 同时被 2 个专栏收录

5 篇文章

订阅专栏

RMQorLCA

2 篇文章

订阅专栏

本文介绍了一种用于构建和查询家族树的算法，通过预处理实现快速查询最近公共祖先及节点信息。利用map存储成员名字，确保名字按字母顺序排列，并通过构建家族树进行父子关系连接。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

/*
 Three operations
 	L     : the same as input, correspond to Preorder traversal.
 	b x   : correspond to ask how many sons does x's father have.
 	c x y : LCA(x, y), there is a trick, if LCA(x, y) is x or y,
		    you should output LCA(x, y)'s father,
		    because in our real life, one can't be one's ancestor.

 About sorting the brothers
 	We can use map to store their names.
	Then their names will be in alphabetical order.
	At last, Scan the map and construct the family tree.
 */
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <map>
using namespace std;
//STL
#define MS(c, a) memset(c, a, sizeof c)
//Syntax
#define Rep(c, a, b) for (int c = (a); c < (b); c++)
//Constant
const int MAXN = 30005 << 1;
//Array
typedef int AI[MAXN];

//FSG
struct Edge{
    int v, next;
};
vector<Edge> E;
AI L;
void G_ini(){
	E.clear(); MS(L, -1);
}
void AE(int u, int v){
	Edge te = {v, L[u]};
	L[u] = E.size();
	E.push_back(te);
}
#define Ere(c, a, L) for (int c = L[a]; c != -1; c = E[c].next)

//RMQ
int  ind;
AI dep, ta, id, st[20];
void TTI(int u, int fa = -1)
{
	if (fa != -1) dep[u] = dep[fa] + 1;
	id[u] = ind;
	ta[ind++] = u;
	for (int i = L[u]; i != -1; i = E[i].next)
		if (E[i].v != fa)
		{
			TTI(E[i].v, u);
			ta[ind++] = u;
		}
}
int low(int x)
{
	int o = 0;
	while ((1 << o) <= x) o++;
	return o;
}
void st_ini()
{
	Rep(i, 0, ind) st[0][i] = ta[i];
	int l = low(ind);
	Rep(i, 1, l) Rep(j, 0, ind)
		if (j + (1 << i) < ind)
		{
			int x = st[i - 1][j],
				y = st[i - 1][j + (1 << (i - 1))];
			st[i][j] = (dep[x] < dep[y])? x: y;
		}
}
int rmq(int x, int y)
{
	x = id[x];
	y = id[y];
	if (x > y) swap(x, y);
	int l = low(y - x + 1) - 1;
	x = st[l][x];
	y = st[l][y - (1 << l) + 1];
	return (dep[x] < dep[y])? x: y;
}

int n, q;
AI fat, now, son;
char buf[MAXN];
string nam[MAXN];
map<string, int> H;
map<string, int>::reverse_iterator Hi; 

void LT(int u, int d = 0)
{
	Rep(i, 0, d) putchar('.');
	printf("%s\n", nam[u].c_str());
	Ere(i, u, L) LT(E[i].v, d + 1);
}

int main()
{
	while (scanf("%d", &n), n)
	{
		MS(fat, -1);
		MS(now, -1);
		MS(son, 0);
		H.clear();
		scanf("%s", buf);
		string ne = buf;
		H[ne] = 1;
		nam[now[0] = 0] = ne;
		Rep(i, 2, n + 1)
		{
			scanf("%s", buf);
			int k = 0;
			while (buf[k] == '.') k++;
			son[fat[i - 1] = now[k - 1]]++; 
			H[ne = buf + k] = i;
			nam[now[k] = i - 1] = ne;
		}
		G_ini();
		for (Hi = H.rbegin(); Hi != H.rend(); Hi++)
		{
			int v = Hi->second - 1, u = fat[v];
			if (Hi->second > 1) AE(u, v);
		}
		MS(dep, 0);
		TTI(ind = 0);
		st_ini();
		scanf("%d", &q);
		while (q--)
		{
			char op[3];
			scanf("%s", op);
			if (op[0] == 'L') LT(0);
			if (op[0] == 'b')
			{
				scanf("%s", buf);
				ne = buf;
				int u = H[ne] - 1;
				if (!u) puts("1");
				else printf("%d\n", son[fat[u]]);
			}
			if (op[0] == 'c')
			{
				scanf("%s", buf);
				ne = buf;
				int u = H[ne] - 1;
				scanf("%s", buf);
				ne = buf;
				int v = H[ne] - 1, p = rmq(u, v);
				if (p == u || p == v) p = fat[p];
				printf("%s\n", nam[p].c_str());
			}
		}
	}
	return 0;
}