[NOI2018]归程 kruskal重构树

[NOI2018]归程

LG传送门

kruskal重构树模板题。

另一篇文章里有关于kruskal重构树更详细的介绍和更板子的题目。

题意懒得说了，这题的关键在于快速找出从查询的点出发能到达的点（即经过海拔高于水位线的边能到达的点）中距离(1)号点最近的距离。

看上去可以kruskal，假设我们把边实现按海拔从大到小排序，考虑我们的重构树的性质：一个小根堆，任意一个点到根节点的路径上的点权单调不升，且这条路径上最浅的高于水位线的点(u)的子树中的所有叶节点就是这个点所能到达的所有点。dijkstra预处理每个点到(1)的最短路，再在重构树上一遍dfs处理出每棵子树中到(1)的最短距离，查询时(u)直接倍增求，这题就做完了。

#include <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>
#include <queue>
#define R register
#define I inline
#define B 1000000
using namespace std;
const int N = 200003, M = 400003, inf = 2e9;
char buf[B], *p1, *p2;
I char gc() { return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, B, stdin), p1 == p2) ? EOF : *p1++; }
I int rd() {
    R int f = 0;
    R char c = gc();
    while (c < 48 || c > 57)
        c = gc();
    while (c > 47 && c < 58)
        f = f * 10 + (c ^ 48), c = gc();
    return f;
}
int s[N], vis[N], dis[N], f[M], o[M], val[M], son[M][2], dep[M], fa[M][20], n, cnt;
struct edge { int u, v, l, a; }e[M];
vector <pair <int, int> > g[N];
priority_queue <pair <int, int> > q;
I int operator < (edge x, edge y) { return x.a > y.a; }
I int min(int x, int y) { return x < y ? x : y; }
I int find(int x) {
    R int r = x, y;
    while (f[r] ^ r)
        r = f[r];
    while (x ^ r)
        y = f[x], f[x] = r, x = y;
    return r;
}
void dfs(int x, int f) {
    dep[x] = dep[f] + 1, fa[x][0] = f;
    for (R int i = 1; i < 20; ++i)
        fa[x][i] = fa[fa[x][i - 1]][i - 1];
    if (x <= n) {
        o[x] = dis[x];
        return ;
    }
    dfs(son[x][0], x), dfs(son[x][1], x), o[x] = min(o[son[x][0]], o[son[x][1]]);
}
I int query(int x, int y) {
    for (R int i = 19; ~i; --i)
        if (dep[x] - (1 << i) > 0 && val[fa[x][i]] > y)
            x = fa[x][i];
    return o[x];
}
int main() {
    R int T = rd(), m, Q, K, S, i, x, y, z, last;
    while (T--) {
        memset(vis, 0, sizeof vis), n = rd(), m = rd(), last = 0;
        for (i = 1; i <= n; ++i)
            g[i].clear();
        for (i = 1; i <= m; ++i) {
            x =rd(), y = rd(), z =rd(), e[i] = (edge){x, y, z, rd()};
            g[x].push_back(make_pair(y, z)), g[y].push_back(make_pair(x, z));
        }
        for (i = 1; i <= n; ++i)
            s[i] = g[i].size(), dis[i] = inf;
        for (i = 1; i < n << 1; ++i)
            f[i] = i;
        dis[1] = 0, q.push(make_pair(0, 1));
        while (!q.empty()) {
            x = q.top().second, q.pop();
            if (vis[x])
                continue;
            vis[x] = 1;
            for (i = 0; i < s[x]; ++i) {
                y = g[x][i].first, z = g[x][i].second;
                if (dis[y] > dis[x] + z)
                    dis[y] = dis[x] + z, q.push(make_pair(-dis[y], y));
            }
        }
        sort(e + 1, e + m + 1), cnt = n;
        for (i = 1; i <= m; ++i) {
            x = find(e[i].u), y = find(e[i].v);
            if (x ^ y) {
                ++cnt, f[x] = f[y] = cnt, val[cnt] = e[i].a;
                son[cnt][0] = x, son[cnt][1] = y;
            }
        }
        dfs(cnt, 0), Q = rd(), K = rd(), S = rd();
        for (i = 1; i <= Q; ++i) {
            x = (rd() + K * last - 1) % n + 1, y = (rd() + K * last) % (S + 1);
            printf("%d
", last = query(x, y));
        }
    }
    return 0;
}