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 + , oo = 1e9 + ;
 
struct Node {
    int u, v, len, high, nxt;
} E[N], G[N << ], Edge[N << ];
struct Node_ {
    int u, dis_;
    inline bool operator < (const Node_ a) const {return dis_ > a.dis_;}
};
 
int head_1[N], head_2[N << ], now;
int dis[N << ];
bool vis[N];
int fa[N << ];
int n, m;
int High[N << ];
int f[N << ][];
int deep[N << ];
 
#define gc getchar()
inline int read() {
    int x = ;
    char c = gc;
    while(c < '' || c > '') c = gc;
    while(c >= '' && c <= '') x = x *  + c - '', 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][] = u, Dfs(G[i].v, u);}
inline int Jump(int X, int H) {for(int i = ; i >= ; 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 = ; i <= ; i ++) for(int j = ; j <= (n *  - ); j ++) f[j][i] = f[f[j][i - ]][i - ];}
 
inline void Dijkstra() {
    for(int i = ; i <= n; i ++) dis[i] = oo;
    for(int i = ; i <= n; i ++) vis[i] = ;
    priority_queue <Node_> Q;
    Q.push((Node_) {, });
    dis[] = ;
    while(!Q.empty()) {
        Node_ topp = Q.top();
        Q.pop();
        if(vis[topp.u]) continue;
        vis[topp.u] = ;
        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 + , E + m + , Cmp);
    for(int i = ; i <= (n << ); i ++) fa[i] = i;
    for(int i = ; i <= (n << ); i ++) head_2[i] = -;
    int cnt = n;
    now = ;
    for(int i = ; i <= m; i ++) {
        if(cnt == n *  - ) 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, , sizeof f);
        n = read(), m = read(); now = ;
        for(int i = ; i <= n; i ++) head_1[i] = -;
        for(int i = ; 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( * n - , ), Pre();
        int Q = read(), K = read(), S = read(), Lastans = ;
        for(; Q; Q --) {
            int X = (read() + K * Lastans - ) % n + , H = (read() + K * Lastans) % (S + );
            Lastans = dis[Jump(X, H)]; cout << Lastans << "\n";
        }
    }
    return ;
}