BZOJ 1787 AHOI 2008 Meet 紧急集合 倍增LCA

标签：bzoj   lca   倍增   ahoi2008   

题目大意：给出一棵树，在上满找三个点，问那个点到这三个点的距离和最短。

思路：可以证明，这个店必然是这三个点之间两个的LCA，然后枚举就可以了。

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 1000010
#define INF 0x3f3f3f3f
using namespace std;

int points,asks;
int head[MAX],total;
int next[MAX],aim[MAX];

int deep[MAX],father[MAX][20];
int length[MAX];

inline void Add(int x,int y)
{
	next[++total] = head[x];
	aim[total] = y;
	head[x] = total;
}

void DFS(int x,int last)
{
	deep[x] = deep[last] + 1;
	length[x] = length[last] + 1; 
	for(int i = head[x]; i; i = next[i]) {
		if(aim[i] == last)	continue;
		father[aim[i]][0] = x;
		DFS(aim[i],x);
	}
}

void SparseTable()
{
	for(int j = 1; j < 20; ++j)
		for(int i = 1; i <= points; ++i)
			father[i][j] = father[father[i][j - 1]][j - 1];
}

inline int GetLCA(int x,int y)
{
	if(deep[x] < deep[y])	swap(x,y);
	for(int i = 19; ~i; --i)
		if(deep[father[x][i]] >= deep[y])
			x = father[x][i];
	if(x == y)	return x;
	for(int i = 19; ~i; --i)
		if(father[x][i] != father[y][i])
			x = father[x][i],y = father[y][i];
	return father[x][0];
}

inline int GetLength(int x,int y,int lca)
{
	return length[x] + length[y] - (length[lca] << 1);
}

int main()
{
	cin >> points >> asks;
	for(int x,y,i = 1; i < points; ++i) {	
		scanf("%d%d",&x,&y);
		Add(x,y),Add(y,x);
	}
	DFS(1,0);
	SparseTable();
	for(int x,y,z,i = 1; i <= asks; ++i) {
		scanf("%d%d%d",&x,&y,&z);
		int lca = GetLCA(x,y),ans = INF,p = 0;
		int temp = GetLength(x,y,lca) + GetLength(z,lca,GetLCA(z,lca));
		if(temp < ans)	ans = temp,p = lca;
		lca = GetLCA(y,z);
		temp = GetLength(y,z,lca) + GetLength(x,lca,GetLCA(x,lca));
		if(temp < ans)	ans = temp,p = lca;
		lca = GetLCA(x,z);
		temp = GetLength(x,z,lca) + GetLength(y,lca,GetLCA(y,lca));
		if(temp < ans)	ans = temp,p = lca;
		printf("%d %d\n",p,ans);
	}
	return 0;
}

BZOJ 1787 AHOI 2008 Meet 紧急集合 倍增LCA

标签：bzoj   lca   倍增   ahoi2008   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/40743709