[NOI2018]归程

嘟嘟嘟


敲，要是会kruskal重构树，这题也太水了吧。
真的，我感觉模板题都比这难。


首先用dijkstra求出点1到每一个点的最短路，然后以\(a\)为关键字把边从大到小排序建出kruskal重构树。这样这棵树就是一个小根堆。
对于每一个询问，从\(v\)倍增往上跳，直到有一个点\(x\)满足\(a[x] > p\)且最小。那么\(x\)的子树中的所有点\(v\)都可以走到，令每个点的权值为1到他的最短路，则答案就是\(x\)子树内权值的最小值。
这个最小值dfs预处理一下就好了。


上午没学kruskal重构树的时候这题上手暴力，50分的\(O(n ^ 2logn)\)暴力dij加上10分树的情况，代码都比这个长……

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2e5 + 5;
const int N = 19;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m, q, K, S, ncnt;
int a[maxn << 1], p[maxn << 1];
struct Node
{
  int x, y, w, a;
  In bool operator < (const Node& oth)const
  {
    return a > oth.a;
  }
}E[maxn << 1];
struct Edge
{
  int nxt, to, w;
}e[maxn << 2];
int head[maxn << 1], ecnt = -1;
In void addEdge(int x, int y, int w)
{
  e[++ecnt] = (Edge){head[x], y, w};
  head[x] = ecnt;
}

In int Find(int x) {return x == p[x] ? x : p[x] = Find(p[x]);}

#define pr pair<int, int>
#define mp make_pair
bool in[maxn];
int dis[maxn];
In void dijkstra(int s)
{
  for(int i = 1; i <= n; ++i) dis[i] = INF, in[i] = 0;
  priority_queue<pr, vector<pr>, greater<pr> > q;
  q.push(mp(dis[s] = 0, s));
  while(!q.empty())
    {
      int now = q.top().second; q.pop();
      if(in[now]) continue;
      in[now] = 1;
      for(int i = head[now], v; ~i; i = e[i].nxt)
	{
	  if(dis[v = e[i].to] > dis[now] + e[i].w)
	    {
	      dis[v] = dis[now] + e[i].w;
	      q.push(mp(dis[v], v));
	    }
	}
    }
}

int dep[maxn << 1], fa[N + 2][maxn << 1], Min[maxn << 1];
In void dfs(int now)
{
  Min[now] = now <= n ? dis[now] : INF;
  for(int i = 1; (1 << i) <= dep[now]; ++i)
    fa[i][now] = fa[i - 1][fa[i - 1][now]];
  for(int i = head[now], v; ~i; i = e[i].nxt)
    {
      dep[v = e[i].to] = dep[now] + 1;
      fa[0][v] = now;
      dfs(v);
      Min[now] = min(Min[now], Min[v]);
    }
}

In int solve(int x, int p)
{
  for(int i = N; i >= 0; --i)
    if(fa[i][x] && a[fa[i][x]] > p) x = fa[i][x];
  return x;
}

In void init()
{
  Mem(head, -1), ecnt = -1;
  Mem(fa, 0), Mem(dep, 0);
  for(int i = 1; i <= n; ++i) p[i] = i;
}

int main()
{
  freopen("return.in", "r", stdin);
  freopen("return.out", "w", stdout);
  int T = read();
  while(T--)
    {
      n = read(), m = read();
      init();
      for(int i = 1; i <= m; ++i)
	{
	  int x = read(), y = read(), w = read(), a = read();
	  addEdge(x, y, w), addEdge(y, x, w);
	  E[i] = (Node){x, y, w, a};
	}
      dijkstra(1);
      Mem(head, -1), ecnt = -1;
      sort(E + 1, E + m + 1); ncnt = n;
      for(int i = 1; i <= m; ++i)
	{
	  int px = Find(E[i].x), py = Find(E[i].y);
	  if(px == py) continue;
	  a[++ncnt] = E[i].a;
	  p[ncnt] = ncnt;
	  addEdge(ncnt, px, 0), addEdge(ncnt, py, 0);
	  p[px] = ncnt, p[py] = ncnt;
	}
      dfs(ncnt);
      q = read(), K = read(), S = read();
      for(int i = 1, ans = 0; i <= q; ++i)
	{
	  int v = (read() + K * ans - 1) % n + 1;
	  int p = (read() + K * ans) % (S + 1);
	  v = solve(v, p);
	  write(ans = Min[v]), enter;
	}
    }
  return 0;
}