hdu 2586 how far away?
Problem Description
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
在线算法：
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
题意：给出一棵树，求出树上某两点之间的（最短）距离。
LCA 模板题
离线算法：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int inf=0x7f7f7f7f;
struct node
{
int next,en,len;
}E[100000];
struct node1
{
int next,en;
}query[100000];
int num1,num2,head1[100000],head2[100000],fa[100000],dis[100000],ans[100000];
int n,m;
bool vis[100000];
void init()
{
num1=num2=0;int i;
memset(head1,-1,sizeof(head1));
memset(head2,-1,sizeof(head2));
memset(vis,false,sizeof(vis));
memset(dis,inf,sizeof(dis));
for(i=0;i<=n;i++)
fa[i]=i;
}
void add1(int st,int en,int len)
{
E[num1].en=en;E[num1].len=len;
E[num1].next=head1[st];head1[st]=num1++;
}
void add2(int st,int en)
{
query[num2].en=en;query[num2].next=head2[st];head2[st]=num2++;
}
int find(int x)
{
if(x==fa[x]) return fa[x];
else return fa[x]=find(fa[x]);
}
void dfs(int x)
{
vis[x]=true;
int i;
for(i=head2[x];i!=-1;i=query[i].next)
{
int v=query[i].en;
if(vis[v])
ans[i>>1]=dis[x]+dis[v]-2*dis[find(v)];
}
for(i=head1[x];i!=-1;i=E[i].next)
{
int v=E[i].en;
if(!vis[v])
{
dis[v]=dis[x]+E[i].len;
dfs(v);
fa[v]=x;
}
}
}
int main()
{
int T,i,j;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
init();
for(i=1;i<=n-1;i++)
{
int st,en,len;
scanf("%d%d%d",&st,&en,&len);
add1(st,en,len);add1(en,st,len);
}
for(i=0;i<m;i++)
{
int st,en;
scanf("%d%d",&st,&en);
add2(st,en);add2(en,st);
}
dis[1]=0;
dfs(1);
for(i=0;i<num2;i+=2)
{
printf("%d\n",ans[i>>1]);
}
}
return 0;
}
在线算法：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
struct node
{
int en,next,len;
}E[80010];
int num,head[50000];
int deep[80010],fa[80010],dis[50000],p[80010][20];//p[i][j]表示i结点的第2^j祖先，dis[i]表示到根的距离
int n,m;
void init()
{
num=0;int i;
memset(head,-1,sizeof(head));
}
void add(int st,int en,int len)
{
E[num].en=en;E[num].len =len;
E[num].next=head[st];head[st]=num++;
}
void dfs(int x,int father,int depth)
{
deep[x]=depth;
fa[x]=father;
int i ,v;
for(i=head[x];i!=-1;i=E[i].next)
{
v=E[i].en;
if(v!=father)
{
dis[v]=dis[x]+E[i].len;
dfs(v,x,depth+1);
}
}
}
void rmq()//求最近公共祖先
{
int i,j;
memset(p,-1,sizeof(p));
for(i=1; i <= n; i++)
p[i][0] = fa[i];
for(j=1; (1<<j) <= n; j++)
{
for(i=1; i <= n; i++)
{
if(p[i][j-1]!=-1)
p[i][j]=p[p[i][j-1]][j-1];//i的第2^j祖先就是i的第2^(j-1)祖先的第2^(j-1)祖先
}
}
}
int query(int st,int en)
{
if(deep[st]<deep[en]) swap(st,en);
int i,j;
for(i=0;(1<<i)<=deep[st];i++);
i--; //使st，en两点的深度相同
for(j=i;j>=0;j--)
if(deep[st]-(1<<j)>=deep[en])
st=p[st][j];
if(st==en) return st;
for(j=i;j>=0;j--)//倍增法，每次向上进深度2^j，找到最近公共祖先的子结点
{
if(p[st][j]!=-1&&p[st][j]!=p[en][j])
{
st=p[st][j];en=p[en][j];
}
}
return fa[st];
}
int main()
{
int T,st,en,len,i,j;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
init();
for(i=1;i<n;i++)
{
scanf("%d%d%d",&st,&en,&len);
add(st,en,len);add(en,st,len);
}
dis[1]=0;dfs(1,-1,0);
rmq();
while(m--)
{
scanf("%d%d",&st,&en);
int ans=query(st,en);
printf("%d\n",dis[st]+dis[en]-2*dis[ans]);
}
}
return 0;
}