luogu 4768

kruskal 重构树
对于一张无向图，我们在进行 kruskal 的过程中
每当合并两个联通块时
新建虚拟节点 t
对于两个联通块的根节点 fau，fav 连无向边
(fau, t)，(fav, t) 其中点 t 的点权为两个联通块当前连边的边权
对于这道题
首先 dijkstra 处理所有点到1号点的最短路
然后按照边的海拔进行降序排序
这样做出重构树之后
显然对于点 u，它的所有子树中的相关的边的海拔(这里已经转化为了虚拟节点的点权)都要大于该点的海拔
这样的话
对于询问二元组 x, h
倍增将 x 调到海拔最低且高于 h 的点处
此时 x 的子树中dis[]的最小值即为此次询问的结果
注意：在进行重构树时
虚拟节点的dis[]每次可以取 min(dis[fau], dis[fav])
这样就相当于dis[t]表示 t 的子树中dis[]的最小值
省去了一遍 dfs

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>

using namespace std;
const int N = 4e5 + 10, oo = 1e9 + 7;

struct Node {
    int u, v, len, high, nxt;
} E[N], G[N << 2], Edge[N << 1];
struct Node_ {
    int u, dis_;
    inline bool operator < (const Node_ a) const {return dis_ > a.dis_;}
};

int head_1[N], head_2[N << 1], now;
int dis[N << 1];
bool vis[N];
int fa[N << 1];
int n, m;
int High[N << 1];
int f[N << 1][30];
int deep[N << 1];

#define gc getchar()
inline int read() {
    int x = 0;
    char c = gc;
    while(c < '0' || c > '9') c = gc;
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
    return x;
}

int Get(int x) {return fa[x] == x ? x : fa[x] = Get(fa[x]);}
inline bool Cmp(Node a, Node b) {return a.high > b.high;}
void Dfs(int u, int fa) {for(int i = head_2[u]; ~ i; i = G[i].nxt) if(G[i].v != fa) f[G[i].v][0] = u, Dfs(G[i].v, u);}
inline int Jump(int X, int H) {for(int i = 20; i >= 0; i --) if(f[X][i] && High[f[X][i]] > H) X = f[X][i];return X;}
inline void Add_Edge(int u, int v, int Len) {Edge[++ now].v = v; Edge[now].len = Len; Edge[now].nxt = head_1[u]; head_1[u] = now;}
inline void Add_G(int u, int v) {G[++ now].v = v; G[now].nxt = head_2[u]; head_2[u] = now;}
inline void Pre() {for(int i = 1; i <= 20; i ++) for(int j = 1; j <= (n * 2 - 1); j ++) f[j][i] = f[f[j][i - 1]][i - 1];}

inline void Dijkstra() {
    for(int i = 1; i <= n; i ++) dis[i] = oo;
    for(int i = 1; i <= n; i ++) vis[i] = 0;
    priority_queue <Node_> Q;
    Q.push((Node_) {1, 0});
    dis[1] = 0;
    while(!Q.empty()) {
        Node_ topp = Q.top();
        Q.pop();
        if(vis[topp.u]) continue;
        vis[topp.u] = 1;
        for(int i = head_1[topp.u]; ~ i; i = Edge[i].nxt)
            if(dis[Edge[i].v] > dis[topp.u] + Edge[i].len) {
                dis[Edge[i].v] = dis[topp.u] + Edge[i].len;
                Q.push((Node_) {Edge[i].v, dis[Edge[i].v]});
            }
    }
}

inline void Kruskal() {
    sort(E + 1, E + m + 1, Cmp);
    for(int i = 1; i <= (n << 1); i ++) fa[i] = i;
    for(int i = 1; i <= (n << 1); i ++) head_2[i] = -1;
    int cnt = n;
    now = 0;
    for(int i = 1; i <= m; i ++) {
        if(cnt == n * 2 - 1) break;
        int u = E[i].u, v = E[i].v, fau = Get(u), fav = Get(v);
        if(fau != fav) {
            fa[fau] = fa[fav] = ++ cnt;
            High[cnt] = E[i].high;
            dis[cnt] = min(dis[fau], dis[fav]);
            Add_G(fau, cnt), Add_G(cnt, fau), Add_G(fav, cnt), Add_G(cnt, fav);
        }
    }
}

int main() {
    for(int T = read(); T; T --) {
        memset(f, 0, sizeof f);
        n = read(), m = read(); now = 0;
        for(int i = 1; i <= n; i ++) head_1[i] = -1;
        for(int i = 1; i <= m; i ++) {
            int u = read(), v = read(), Len = read(), high = read();
            Add_Edge(u, v, Len), Add_Edge(v, u, Len);
            E[i] = (Node) {u, v, Len, high};
        }
        Dijkstra(), Kruskal(), Dfs(2 * n - 1, 0), Pre();
        int Q = read(), K = read(), S = read(), Lastans = 0;
        for(; Q; Q --) {
            int X = (read() + K * Lastans - 1) % n + 1, H = (read() + K * Lastans) % (S + 1);
            Lastans = dis[Jump(X, H)]; cout << Lastans << "\n";
        }
    }
    return 0;
}