#294 (div.2) E.A and B and Lecture Rooms

标签：lca

1.题目描述：点击打开链接

2.解题思路：本题让我长见识了。也学到了很多新的知识：LCA，多级祖先算法。如果只是单纯地将无根树转化为有根树，找到u,v的中点，再用BFS计算中线上结点的个数，那么最终会导致TLE。本题的高效算法如下：

首先求出以1为根的树的所有结点的总个数，保存在num数组中，再利用LCA算法求出u,v的公共祖先，设为LCA。找到u,v结点的中点mid.此时规定deep[u]>deep[v]，不满足就交换两数。那么分两种情况讨论：(i)如果d(u,LCA)==d(v,LCA)（d表示距离），那么答案是n-num[u]-num[v];(ii)如果d(u,LCA)!=d(v,LCA)，设偏向u的中点是midU(mid--即可)，那么答案是num[mid]-num[midU]。

3.代码：

#define _CRT_SECURE_NO_WARNINGS 
#include<iostream>
#include<algorithm>
#include<string>
#include<sstream>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<functional>
using namespace std;

#define N 101010+10
int num[N];
int head[N];//链表表头，head[u]表示与u相连的边序号
int tot;//边序号
int deep[N];//结点深度
int p[N][30];
struct Node
{
	int to, next;
}edge[N << 1];//存放所有的边

void addedge(int from, int to)
{
	edge[tot].to = to;
	edge[tot].next = head[from];
	head[from] = tot++;
}
void dfs(int u, int fa)//无根树转化为1为根的有根树
{
	for (int i = head[u]; ~i; i = edge[i].next)
	{
		int v = edge[i].to;
		if (v == fa)continue;
		deep[v] = deep[u] + 1;
		p[v][0] = u;//直接祖先
		dfs(v, u);
	}
}
int lca(int a, int b)//求最近公共祖先
{
	if (deep[a] < deep[b])
		swap(a, b);
	int d = deep[a] - deep[b];
	for (int i = 0; i < 30;i++)
	if (d&(1 << i))
		a = p[a][i];
	if (a == b)
		return a;
	for (int i = 29; i >= 0;i--)
	if (p[a][i] != p[b][i])
	{
		a = p[a][i];
		b = p[b][i];
	}
	return p[a][0];
}

int dfs2(int u, int fa)//计算u为根的子树的结点数
{
	num[u] = 1;
	for (int i = head[u]; ~i; i = edge[i].next)
	{
		int v = edge[i].to;
		if (v == fa)continue;
		num[u] += dfs2(v, u);
	}
	return num[u];
}
int main()
{
	freopen("test.txt", "r", stdin);
	int n;
	while (scanf("%d", &n) != EOF)
	{
		int u, v;
		memset(head, -1, sizeof(head));
		tot = 0;
		for (int i = 0; i < n - 1; i++)
		{
			scanf("%d%d", &u, &v);
			addedge(u, v);
			addedge(v, u);
		}
		dfs(1, -1);//建树
		memset(num, 0, sizeof(num));
		deep[1] = 0;
		dfs2(1, -1);//统计每个结点的结点数
		for (int j = 1; j < 30;j++)
		for (int i = 1; i <= n; i++)
			p[i][j] = p[p[i][j - 1]][j - 1];//i的第j层祖先等于i的第j-1层祖先的祖先
		int m;
		scanf("%d", &m);
		while (m--)
		{
			scanf("%d%d", &u, &v);
			if (u == v)
			{
				printf("%d\n", n);
				continue;
			}
			int LCA = lca(u, v);
			int d1 = deep[u] - deep[LCA];
			int d2 = deep[v] - deep[LCA];
			if (d1 != d2)
			{
				if (abs(d1 - d2) & 1)//距离差是奇数，输出0
					printf("0\n");
				else
				{
					if (deep[u] < deep[v])
						swap(u, v);
					int dist = (d1 + d2) / 2;
					int uu = u;
					for (int k = 0; k < 30;k++)
					if (dist&(1 << k))
						uu = p[uu][k];
					dist--;
					int vv = u;
					for (int k = 0; k < 30;k++)
					if (dist&(1 << k))
						vv = p[vv][k];
					printf("%d\n", num[uu] - num[vv]);
				}
			}
			else
			{
				int uu = u;
				int dist = d1 - 1;
				for (int k = 0; k < 30;k++)
				if (dist&(1 << k))
					uu = p[uu][k];
				int vv = v;
				for (int k = 0; k < 30;k++)
				if (dist&(1 << k))
					vv = p[vv][k];
				printf("%d\n", n - num[vv] - num[uu]);
			}
		}
	}
	return 0;
}

#294 (div.2) E.A and B and Lecture Rooms

标签：lca

原文地址：http://blog.csdn.net/u014800748/article/details/44007129