归程p4768

link
如果把所有互相之间的边没有积水的点看成一个个的块，显然块内部可以开车直接到达，也就是说块内部所有点的答案是一样的；
我们可以先用dij把每一个点到1号点的最短路预处理出来，建一棵kruskal重构树，然后在dfs时把这一个子树内的最小答案挂在根节点上；
最后我们倍增找到出发节点所在块的根节点（最浅的那一个），输出其答案即可；

#include<bits/stdc++.h>
#define m(a) memset(a, 0, sizeof(a))
using namespace std;
const int N = (2e5 + 5) * 2;
const int M = 4e5 + 5;
int T, n, m, q, k, s, dis[N], vis[N], fa[N], f[N][22], dq[N];
vector<pair<int, int> > mp[N];
vector<int> tr[N];
struct edge{
	int u, v, l, a;
}e[M];
struct node{
	int id, w;
	friend bool operator <(node a, node b) {
		return a.w > b.w;
	}
};
bool cmp(edge x, edge y){
	return x.a > y.a;
}
int find(int x){
	if(fa[x] == x) return x;
	return fa[x] = find(fa[x]);
}
void dij(){
	priority_queue<node> q;
	for(int i = 1; i <= n; i++) dis[i] = 0x3f3f3f3f;
	dis[1] = 0;
	q.push({1, 0});
	while(!q.empty()){
		int u = q.top().id;
		q.pop();
		if(vis[u]) continue;
		vis[u] = 1;
		for(auto tmp : mp[u]){
			int v = tmp.first;
			int w = tmp.second;
			if(dis[v] > dis[u] + w){
				dis[v] = dis[u] + w;
				q.push({v, dis[v]});
			}
		}
	}	
}
void kruskal(){
	sort(e + 1, e + m + 1, cmp);
	for(int i = 1; i <= n; i++) fa[i] = i, dq[i] = 0;
	for(int i = 1; i <= m; i++){
		int f1 = find(e[i].u);
		int f2 = find(e[i].v);
		if(f1 == f2) continue;
		fa[f1] = fa[f2] = fa[++n] = n;
		dq[n] = e[i].a; 
		dis[n] = 0x3f3f3f3f;
		tr[n].push_back(f1); tr[n].push_back(f2);
	}
}
void dfs(int u, int father){
	f[u][0] = father;
	for(int j = 1; j <= 20; j++) f[u][j] = f[f[u][j - 1]][j - 1];
	for(auto v : tr[u]){
		dfs(v, u);
		dis[u] = min(dis[u], dis[v]);
	}
}
int main(){
	cin >> T;
	while(T--){
		int lastans = 0;
		m(e); m(tr); m(vis); m(dis); m(fa); m(f); m(dq);
		cin >> n >> m;
		int cnt = n;
		for(int i = 1; i <= n * 2; i++) mp[i].clear(), tr[i].clear();
		for(int i = 1; i <= m; i++){
			cin >> e[i].u >> e[i].v >> e[i].l >> e[i].a;
			mp[e[i].u].push_back({e[i].v, e[i].l});
			mp[e[i].v].push_back({e[i].u, e[i].l});
		}
		cin >> q >> k >> s;
		dij();
		kruskal();
		dfs(n, n);
		for(int i = 1; i <= q; i++){
			int v0, p0;
			cin >> v0 >> p0;
			int c = (v0 + k * lastans - 1) % cnt + 1;
			int p = (p0 + k * lastans) % (s + 1);
			for(int j = 20; j >= 0; j--){
				if(dq[f[c][j]] > p){
					c = f[c][j];
				}	
			}	
			lastans = dis[c];
			cout << lastans << endl;
		}
	}
}