atcoder-abc-070D

       

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≤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)

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

题意：
一颗无向树，有n-1条边，求x经过k到达y的最短路径
思路：
转换成从k点到x和y两点的最短路径
注意：
不能用邻接矩阵储存树，数据量过大，要用邻接表；

树为无向树；

        #include <iostream>

        #include <cstdio>
        #include <vector>
        #include <algorithm>
        #include <cstring>
        #include <queue>
        using namespace std;
        const int N=1e5+50;
        const long long INF=1e18;
        int n;
        int k,p;
        long long dis[N];
        int vis[N];
        struct Edge
        {
            int y;
            int w;
        };
        vector<Edge> v[N];
        void Init()
        {
            for(int i=0; i<=n; i++)
                v[i].clear();
        }
        void spfa(int s)
        {
            queue<int> p;
            p.push(s);
            for(int i=0; i<=n; i++) dis[i]=INF;
            dis[s]=0;
            memset(vis,0,sizeof(vis));
            vis[s]=1;
            while(!p.empty())
            {
                int u=p.front();
                p.pop();
                vis[u]=0;
                for(int i=0; i<v[u].size(); i++)
                {
                    int w=v[u][i].w;
                    int y=v[u][i].y;
                    if(dis[y]>dis[u]+w)
                    {
                        dis[y]=dis[u]+w;
                       if(!vis[y])
                       {
                           p.push(y);
                           vis[y]=1;
                       }

                    }
                }
            }
        }
        int main()
        {
            scanf("%d",&n);
            Init();
            for(int i=0; i<n-1; i++)
            {
                int x,y,w;
                scanf("%d%d%d",&x,&y,&w);
                Edge now;
                now.y=y;
                now.w=w;
                v[x].push_back(now);
                now.y=x;
                v[y].push_back(now);
            }
            scanf("%d%d",&p,&k);
            spfa(k);
            for(int i=0; i<p; i++)
            {
                int x,y;
                scanf("%d%d",&x,&y);
                cout<<dis[x]+dis[y]<<endl;
            }
            return 0;
        }