tarjanLCA求树上任意两点间的距离。。。。
任意选一个点当根节点，dist[a,b]= dist[a] +dist[b] -2* lca[a,b]
How far away ？
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4254 Accepted Submission(s): 1622
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.
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
Source
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#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=44000;
struct Edge
{
int to,next,distan;
}edge[2*maxn],edge2[maxn];
int Adj[maxn],Adj2[maxn],Size,Size2;
void init_1()
{
Size=0;
memset(Adj,-1,sizeof(Adj));
}
void init_2()
{
Size2=0;
memset(Adj2,-1,sizeof(Adj2));
}
void Add_Edge(int u,int v,int c)
{
edge[Size].to=v;
edge[Size].next=Adj[u];
edge[Size].distan=c;
Adj[u]=Size++;
}
void Add_Edge2(int u,int v,int id)
{
edge2[Size2].to=v;
edge2[Size2].next=Adj2[u];
edge2[Size2].distan=id;
Adj2[u]=Size2++;
}
int n,m;
int fa[maxn];
int Find(int x)
{
if(x==fa[x]) return x;
return fa[x]=Find(fa[x]);
}
bool vis[maxn];
int Distan[maxn],ans[maxn];
void LCA(int u,int father)
{
for(int i=Adj[u];~i;i=edge[i].next)
{
int v=edge[i].to;
if(v==father) continue;
Distan[v]=Distan[u]+edge[i].distan;
LCA(v,u);
fa[v]=u;
}
vis[u]=true;
for(int i=Adj2[u];~i;i=edge2[i].next)
{
int v=edge2[i].to;
int id=edge2[i].distan;
if(vis[v])
{
ans[id]=Distan[u]+Distan[v]-2*Distan[Find(v)];
}
}
}
int main()
{
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
init_1();init_2();
scanf("%d%d",&n,&m);
fa[0]=0;fa[1]=1;fa[2]=2;
for(int i=0;i<n-1;i++)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
Add_Edge(a,b,c);
Add_Edge(b,a,c);
fa[i+3]=i+3;
}
for(int i=0;i<m;i++)
{
int a,b;
scanf("%d%d",&a,&b);
Add_Edge2(a,b,i);
Add_Edge2(b,a,i);
}
memset(Distan,0,sizeof(Distan));
memset(ans,0,sizeof(ans));
memset(vis,false,sizeof(vis));
LCA(1,1);
for(int i=0;i<m;i++)
printf("%d\n",ans[i]);
}
return 0;
}