链接:
How far away ？
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11204 Accepted Submission(s): 4079
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
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
25
100
100
一个村子里有n个房子，这n个房子用n-1条路连接起来，接下了有m次询问，每次询问两个房子a,b之间的距离是多少。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long Ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-;
const int inf =0x7f7f7f7f;
const double pi=acos(-);
const int maxn=; struct Edge{
int to,w;
Edge(int a,int b):to(a),w(b){};
Edge(){};
}e[maxn+]; struct node{
int to,id;
}; vector<Edge> G[maxn+];
vector<node> mp[maxn+];
int vis[maxn+],query[maxn+][],par[maxn+],dis[maxn+]; int findr(int u)
{
if(par[u]!=u)
par[u]=findr(par[u]);
return par[u];
} void unite(int u,int v)
{
int ru=findr(u);
int rv=findr(v);
if(ru!=rv) par[rv]=ru;
} void tarjan(int u,int val)
{
vis[u]=;
dis[u]=val; for(int i=;i<mp[u].size();i++)
{
int v=mp[u][i].to;
int id=mp[u][i].id;
if(vis[v])
query[id][]=findr(v);
} for(int i=;i<G[u].size();i++)
{
int v=G[u][i].to;
if(vis[v]) continue;
tarjan(v,val+G[u][i].w);
unite(u,v);
} } int main()
{
int cas,n,op,u,v,w;
scanf("%d",&cas);
while(cas--)
{
scanf("%d %d",&n,&op); for(int i=;i<=n;i++)
{
G[i].clear();
mp[i].clear();
vis[i]=;
par[i]=i;
} for(int i=;i<=n-;i++)
{
scanf("%d %d %d",&u,&v,&w);
G[u].push_back(Edge(v,w));
G[v].push_back(Edge(u,w));
} for(int i=;i<=op;i++)
{
scanf("%d %d",&u,&v);
query[i][]=u;
query[i][]=v;
mp[u].push_back((node){v,i});
mp[v].push_back((node){u,i});
} tarjan(,); for(int i=;i<=op;i++)
{
int u=query[i][];
int v=query[i][];
int r=query[i][];
printf("%d\n",dis[u]+dis[v]-*dis[r]);
}
}
return ;
}
分析：用的lca
1.因为有n-1条边，任意两点之间可达，那么就是一棵树；
2，假设当前是询问u和v两个点，他们的LCA是r,dis[]数组代表从根到点的距离，那么u和v两点之间的距离就是dis[u]+dis[v]-2*dis[r]；所以只需要用Tarjan求一求询问的点对的LCA并且记录一下每个点到根（随便哪个点做根都可以）的距离就可以了