HUSTOJ：Transit Tree Path

问题 D: 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.
Constraints
3≤N≤105
1≤ai,bi≤N(1≤i≤N−1)
1≤ci≤109(1≤i≤N−1)
The given graph is a tree.
1≤Q≤105
1≤K≤N
1≤xj,yj≤N(1≤j≤Q)
xj≠yj(1≤j≤Q)
xj≠K,yj≠K(1≤j≤Q)

输入

Input is given from Standard Input in the following format:
N  
a1 b1 c1  
:  
aN−1 bN−1 cN−1
Q K
x1 y1
:  
xQ yQ

 

输出

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.

样例输入

5
1 2 1
1 3 1
2 4 1
3 5 1
3 1
2 4
2 3
4 5

样例输出

3
2
4

提示

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

---

　　题意：求两个点经过k点之后的最短距离

　　HUST得训练赛上遇到得题目，直接dfs求出k点到各个点的最短距离就可以，输出的时候输出到两点距离的和就ok

　　附上代码（已AC）：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAX 100005
#define ll long long

struct Edge{
    ll to;
    ll next;
    ll cost;
};

Edge edge[MAX*];
ll head[MAX];
ll dis[MAX];
void DFS(ll u,ll v,ll w)
{
    dis[u]=w;
    for(ll i=head[u];i!=-;i=edge[i].next)
    {
        ll to=edge[i].to;
        if(to==v)
            continue;
        DFS(to,u,w+edge[i].cost);
    }
    return ;
}
int main()
{
    ll n,u,v,w,q,k;
    scanf("%lld",&n);
    ll cnt=;
    memset(head,-,sizeof(head));
    for(ll i=;i<=n-;i++)
    {
        scanf("%lld%lld%lld",&u,&v,&w);
        edge[cnt].to=v;
        edge[cnt].cost=w;
        edge[cnt].next=head[u];
        head[u]=cnt++;
        edge[cnt].to=u;
        edge[cnt].cost=w;
        edge[cnt].next=head[v];
        head[v]=cnt++;
    }
    scanf("%lld%lld",&q,&k);
    DFS(k,-,);
    while(q--)
    {
        scanf("%lld%lld",&u,&v);
        printf("%lld\n",dis[u]+dis[v]);
    }
    return ;
}