AT_abc070_D .Transit Tree Path

题目链接：D - Transit Tree Path

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题目大意：

给定一个 n个点 构成为树的图，

在经过 点k 的情况下，m个询问，求两个点的最短距离

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思路：

一定经过点k ，可以看成从k点出发到各点的距离

结果就是 两个点分别到点 k 的距离之和

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代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<bitset>
#include<tuple>
#define inf 72340172838076673
#define int long long
#define endl '\n'
#define F first
#define S second
#define  mst(a,x) memset(a,x,sizeof (a))
using namespace std;
typedef pair<int, int> pii;

const int N = 200086, mod = 998244353;

int n, m, k;
int h[N], ne[N], e[N], w[N], idx;
bool st[N];
vector<int> dis(N, inf);
priority_queue<pii, vector<pii>, greater<pii >> q;

void add(int a, int b, int c) {
    w[idx] = c;
    e[idx] = b;
    ne[idx] = h[a];
    h[a] = idx++;
}

void dij() {

    dis[k] = 0;
    q.push({dis[k], k});
    
    while(q.size()) {
        auto [d, u] = q.top();
        q.pop();
        
        if(st[u]) continue;
        st[u] = true;
        
        for(int i = h[u]; ~i; i = ne[i]) {
            int j = e[i];
            if(dis[j]>d + w[i]) {
                dis[j] = d + w[i];
                q.push({dis[j], j});
            }
        }
    }
}

void solve() {

    mst(h, -1);
    cin >> n;
    for (int i = 1; i < n; i++) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c), add(b, a, c);
    }
    cin >> m >> k;
    
    dij();
    
    while (m--) {
        int a, b;
        cin >> a >> b;
        cout << dis[a] + dis[b] << endl;
    }
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    
    int T = 1;
// cin >> T;
    while (T--) solve();
    
    return 0;
}