「2017 山东三轮集训 Day7」Easy

一棵带边权的树,多次询问 $x$ 到编号为 $[l,r]$ 的点最短距离是多少

$n leq 100000$

sol:

动态点分治,每层重心维护到所有点的距离

查询的时候在管辖这个点的 log 层线段树里查就可以了

因为这样每一层的答案只会漏而不会错,所以正确性有保障

不会写点分治了...orz

#include <bits/stdc++.h>
#define LL long long
#define rep(i, s, t) for (register int i = (s), i##end = (t); i <= i##end; ++i)
#define dwn(i, s, t) for (register int i = (s), i##end = (t); i >= i##end; --i)
using namespace std;
inline int read() {
    int x = 0, f = 1;
    char ch;
    for (ch = getchar(); !isdigit(ch); ch = getchar())
        if (ch == '-')
            f = -f;
    for (; isdigit(ch); ch = getchar()) x = 10 * x + ch - '0';
    return x * f;
}
const int maxn = 200010, inf = 1e9;
int n, m;
int first[maxn], to[maxn << 1], nx[maxn << 1], val[maxn], cnt;
inline void add(int u, int v, int w) {
    to[++cnt] = v;
    val[cnt] = w;
    nx[cnt] = first[u];
    first[u] = cnt;
}
int sig, root, size[maxn], pa[maxn], f[maxn], vis[maxn];
inline void findroot(int x, int fa) {
    size[x] = 1;
    f[x] = 0;
    for (int i = first[x]; i; i = nx[i]) {
        if ((to[i] == fa) || (vis[to[i]]))
            continue;
        findroot(to[i], x);
        size[x] += size[to[i]];
        f[x] = max(f[x], size[to[i]]);
    }
    f[x] = max(f[x], sig - size[x]);
    if (f[x] < f[root])
        root = x;
}
inline void build(int x) {
    vis[x] = 1;
    for (int i = first[x]; i; i = nx[i]) {
        if (vis[to[i]])
            continue;
        root = 0;
        sig = size[to[i]];
        findroot(to[i], x);
        pa[root] = x;
        build(root);
    }
}
namespace sp_lca {
int size[maxn], dep[maxn], fa[maxn], bl[maxn], dp[maxn];
inline void dfs1(int x) {
    size[x] = 1;
    for (int i = first[x]; i; i = nx[i]) {
        if (to[i] == fa[x])
            continue;
        fa[to[i]] = x;
        dep[to[i]] = dep[x] + 1;
        dp[to[i]] = dp[x] + val[i];
        dfs1(to[i]);
        size[x] += size[to[i]];
    }
}
inline void dfs2(int x, int col) {
    int k = 0;
    bl[x] = col;
    for (int i = first[x]; i; i = nx[i])
        if (to[i] != fa[x] && size[to[i]] > size[k])
            k = to[i];
    if (!k)
        return;
    dfs2(k, col);
    for (int i = first[x]; i; i = nx[i])
        if (to[i] != fa[x] && to[i] != k)
            dfs2(to[i], to[i]);
}
inline void init() {
    dfs1(1);
    dfs2(1, 1);
}
inline int lca(int x, int y) {
    while (bl[x] != bl[y]) {
        if (dep[bl[x]] < dep[bl[y]])
            swap(x, y);
        x = fa[bl[x]];
    }
    return dep[x] < dep[y] ? x : y;
}
}  // namespace sp_lca
inline int dis(int x, int y) { return sp_lca::dp[x] + sp_lca::dp[y] - (sp_lca::dp[sp_lca::lca(x, y)] << 1); }
int rt[maxn], ls[maxn << 6], rs[maxn << 6], va[maxn << 6], ToT;
inline void Insert(int &x, int l, int r, int pos, int v) {
    if (!x)
        x = ++ToT, va[x] = inf;
    va[x] = min(va[x], v);
    if (l == r)
        return;
    int mid = (l + r) >> 1;
    if (pos <= mid)
        Insert(ls[x], l, mid, pos, v);
    else
        Insert(rs[x], mid + 1, r, pos, v);
}
inline int query(int x, int l, int r, int L, int R) {
    if (!x)
        return inf;
    if (L <= l && r <= R)
        return va[x];
    int mid = (l + r) >> 1, res = inf;
    if (L <= mid)
        res = min(res, query(ls[x], l, mid, L, R));
    if (R > mid)
        res = min(res, query(rs[x], mid + 1, r, L, R));
    return res;
}
int main() {
    // freopen("a.in","r",stdin);
    // freopen("a.out","w",stdout);
    n = read();
    rep(i, 2, n) {
        int u = read(), v = read(), w = read();
        add(u, v, w);
        add(v, u, w);
    }
    f[0] = inf;
    root = 0;
    sig = n;
    findroot(1, 0);
    build(root);
    sp_lca ::init();
    rep(i, 1, n) for (int x = i; x; x = pa[x]) Insert(rt[x], 1, n, i, dis(x, i));
    // rep(i, 1, n) cout << pa[i] << endl;
    int q = read();
    while (q--) {
        int u = read(), v = read(), x = read();
        int cur = x, ans = inf;
        while (cur) {
            ans = min(ans, query(rt[cur], 1, n, u, v) + dis(cur, x));
            cur = pa[cur];
        }
        cout << ans << endl;
    }
}
View Code
原文地址:https://www.cnblogs.com/Kong-Ruo/p/10510795.html