AtCoder Beginner Contest 070 Transit Tree Path

You are given a tree with N vertices.
Here, a tree is a kind of graph, and more specifically, a connected undirected graph with N−1 edges, where N is the number of its vertices.
The i-th edge (1≤i≤N−1) connects Vertices ai and bi, and has a length of ci.

You are also given Q queries and an integer K. In the j-th query (1≤j≤Q):

• find the length of the shortest path from Vertex xj and Vertex yj via Vertex K.

Input

Input is given from Standard Input in the following format:

N  a1 b1 c1  :  aN−1 bN−1 cN−1Q Kx1 y1:  xQ yQ

Output

Print the responses to the queries in Q lines.
In the j-th line j(1≤j≤Q), print the response to the j-th query.

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Sample Input 1

51 2 11 3 12 4 13 5 13 12 42 34 5

Sample Output 1

324

The shortest paths for the three queries are as follows:

• Query 1: Vertex 2 → Vertex 1 → Vertex 2 → Vertex 4 : Length 1+1+1=3
• Query 2: Vertex 2 → Vertex 1 → Vertex 3 : Length 1+1=2
• Query 3: Vertex 4 → Vertex 2 → Vertex 1 → Vertex 3 → Vertex 5 : Length 1+1+1+1=4

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Sample Input 2

71 2 11 3 31 4 51 5 71 6 91 7 113 21 34 56 7

Sample Output 2

51422

The path for each query must pass Vertex K=2.

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Sample Input 3

101 2 10000000002 3 10000000003 4 10000000004 5 10000000005 6 10000000006 7 10000000007 8 10000000008 9 10000000009 10 10000000001 19 10

Sample Output 3

17000000000

题意：给一个无向图连通两点的经过k点的最短的路径是多少；

n个点n-1条边一定是棵树，只要dfs以k为根的树就行了..

#include<cstdio>#include<iostream>#include<string>#include<cstring>#include<vector>#include<algorithm>using namespace std;typedef long long ll;typedef vector< pair<ll,ll> > Root;Root root[1000005];ll L[1000005];void dfs(ll r,ll fa,ll sum){    L[r]=sum;    for(int i=0;i<root[r].size();i++)    {        if(root[r][i].first==fa) continue;        dfs(root[r][i].first,r,root[r][i].second+sum);    }}int main(void){    int n;    cin>>n;    for(int i=1;i<n;i++)    {        ll a,b,val;        cin>>a>>b>>val;        root[a].push_back(make_pair(b,val));        root[b].push_back(make_pair(a,val));    }    ll que,k;    cin>>que>>k;    dfs(k,-1,0);    while(que--)    {        ll a,b;        cin>>a>>b;        cout<<L[a]+L[b]<<endl;    }    return 0;}