【HDU - 2586】How far away ？（LCA最近公共祖先）

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.

Input

First line is a single integer T(T<=10), indicating the number of test cases. 
  For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n. 
  Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

Output

For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.

Sample Input

2
3 2
1 2 10
3 1 15
1 2
2 3

2 2
1 2 100
1 2
2 1

Sample Output

10
25
100
100

思路：

問一棵樹上，某兩個結點的距離，其實可以用最近公共祖先來解，假設根節點爲D，查詢A、B兩點，他們的最近公共祖先爲C，這道題目求解，就是用這個公式|AB|=|AD|+|BD|-2*|CD|。

這裏用的是離線的tarjan算法。（博客）

ac代碼：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<string.h>
#define mod (1000000007)
#define ll long long
#define ull unsigned long long
using namespace std;
struct node{
	int to;
	int w;
	int nex;
}side[1010101];
struct nod{
	int id,x,y,w;
}que[10101];
int head[101010];
int vis[101010];
int dis[101001];
int cnt=0,tot=0;
int f[50010];
int n,m;
int getf(int x)
{
	if(f[x]!=x)
	f[x]=getf(f[x]);
	return f[x];
}
void init()
{
	for(int i=0;i<50000;i++)
	f[i]=i;
	memset(head,-1,sizeof(head));
	memset(vis,0,sizeof(vis));
	cnt=0;
	tot=0;
}
void add(int x,int y,int w)
{
	side[cnt].to=y;
	side[cnt].nex=head[x];
	side[cnt].w=w;
	head[x]=cnt++;
}
void lca(int x,int ff)
{
	for(int i=head[x];i!=-1;i=side[i].nex)
	{
		int ty=side[i].to;
		if(ty==ff)
		continue;
		dis[ty]=dis[x]+side[i].w;//x不是ff 
		lca(ty,x);
	}
	for(int i=0;i<m;i++)
	{
		if(que[i].x==x)
		{
			if(vis[que[i].y])
			{
				que[i].w=getf(que[i].y);
			}
		}
		else if(que[i].y==x)
		{
			if(vis[que[i].x])
			{
				que[i].w=getf(que[i].x);
			}
		}
	}
	vis[x]=1;
	f[x]=ff;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		init();
		
		scanf("%d%d",&n,&m);
		for(int i=0;i<n-1;i++)
		{
			int u,v,w;
			scanf("%d%d%d",&u,&v,&w);
			add(u,v,w);
			add(v,u,w);
		}
		for(int i=0;i<m;i++)
		{
			scanf("%d%d",&que[i].x,&que[i].y);//也可以開個鄰接表存，並且記錄下id,最後查詢時開一個一維數組，把找到的公共祖先存放到一維數組中這兩個點的id的位置，最後直接遍歷一遍數組就好 
			que[i].id=i;
		}
		lca(1,1);
		for(int i=0;i<m;i++)
		{
			printf("%d\n",dis[que[i].x]+dis[que[i].y]-2*dis[que[i].w]);
		}
	} 
	return 0;
}