AtCoder Beginner Contest 070-D

D - Transit Tree Path

Time limit : 2sec / Memory limit : 256MB

Score : 400 points
Problem Statement

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≤1051≤ai,bi≤N(1≤i≤N−1)1≤ci≤109(1≤i≤N−1)The given graph is a tree.1≤Q≤1051≤K≤N1≤xj,yj≤N(1≤j≤Q)xj≠yj(1≤j≤Q)xj≠K,yj≠K(1≤j≤Q)

Input

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

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.
Sample Input 1

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

Sample Output 1

3
2
4

The shortest paths for the three queries are as follows:

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

Sample Input 2

7
1 2 1
1 3 3
1 4 5
1 5 7
1 6 9
1 7 11
3 2
1 3
4 5
6 7

Sample Output 2

5
14
22

The path for each query must pass Vertex K=2.
Sample Input 3

10
1 2 1000000000
2 3 1000000000
3 4 1000000000
4 5 1000000000
5 6 1000000000
6 7 1000000000
7 8 1000000000
8 9 1000000000
9 10 1000000000
1 1
9 10

Sample Output 3

17000000000

题目大意：给出一个根，问根到两点的距离之和。
解题思路：因为是树，用dfs就行了。

    #include<iostream>    #include<cstdio>    #include<vector>    #include<cmath>    #include<cstring>    #include<fstream>    #include<algorithm>    using namespace std;    typedef long long LL;    const int INF=0x3f3f3f3f;    const int MAXN=1e5+5;    int tot,head[MAXN];    LL dist[MAXN];    bool vis[MAXN];    struct Edge    {        int from,to,nxt;        LL cost;    }e[MAXN*2];    void addedge(int u,int v,int w)    {        e[tot].from=u;        e[tot].to=v;        e[tot].cost=w;        e[tot].nxt=head[u];        head[u]=tot++;    }    void dfs(int u,int fa)    {        for(int i=head[u];i!=-1;i=e[i].nxt)        {            int to=e[i].to;            if(to==fa) continue;            dist[to]=dist[u]+e[i].cost;            dfs(to,u);        }    }    int main()    {        int n;        while(scanf("%d",&n)!=EOF)        {            tot=0;            memset(head,-1,sizeof(head));            memset(dist,0,sizeof(dist));            int u,v;            LL w;            for(int i=1;i<=n-1;i++)            {                scanf("%d%d%lld",&u,&v,&w);                addedge(u,v,w);                addedge(v,u,w);            }            int q,k;            scanf("%d%d",&q,&k);            dfs(k,-1);    //        for(int i=1;i<=n;i++)    //            cout<<dist[i]<<" ";    //        cout<<endl;            int x,y;            for(int i=1;i<=q;i++)            {                scanf("%d%d",&x,&y);                printf("%lld\n",dist[x]+dist[y]);            }        }        return 0;    }