hdu5993/2016icpc青岛L

zz:https://www.cnblogs.com/ytytzzz/p/9674661.html

题意：
给一棵树，每次询问删掉两条边，问剩下的三棵树的最大直径
点10W，询问10W，询问相互独立

Solution：
考虑线段树/倍增维护树的直径
考虑一个点集的区间 [l, r]
而我们知道了有 l <= k < r，
且知道 [l, k] 和 [k + 1, r] 两个区间的最长链的端点及长度
假设两个区间的直径端点分别为 (l1, r1) 和 (l2, r2)
那么 [l, r] 这个区间的直径长度为
dis(l1, r1) dis(l1, l1) dis(l1, r2)
dis(r1, l2) dis(r1, r2) dis(l2, r2)
六个值中的最大值
本题因为操作子树，所以我们维护dfs序的区间最长链即可

#include <stdio.h>
#include <algorithm>

using namespace std;

const int N = 2e5 + 5;

int T, n, m;

int len, head[N], ST[20][N];

struct edge{int u, v, w;}ee[N];

int cnt, fa[N], log_2[N], st[N], en[N], dfn[N], dis[N], dep[N], pos[N];

struct edges{int to, next, cost;}e[N];

inline void add(int u, int v, int w) {
    e[++ len] = (edges){v, head[u], w}, head[u] = len;
    e[++ len] = (edges){u, head[v], w}, head[v] = len;
}

inline void dfs1(int u) {
    st[u] = ++ cnt, dfn[cnt] = u;
    for (int v, i = head[u]; i; i = e[i].next) {
        v = e[i].to;
        if (v == fa[u]) continue;
        fa[v] = u, dep[v] = dep[u] + 1;
        dis[v] = dis[u] + e[i].cost, dfs1(v);
    }
    en[u] = cnt;
}

inline void dfs2(int u) {
    dfn[++ cnt] = u, pos[u] = cnt;
    for (int v, i = head[u]; i; i = e[i].next) {
        v = e[i].to;
        if (v == fa[u]) continue;
        dfs2(v), dfn[++ cnt] = u;
    }
}

int mmin(int x, int y) {
    if (dep[x] < dep[y]) return x;
    return y;
}

inline int lca(int u, int v) {
    static int w;
    if (pos[u] > pos[v]) swap(u, v);
    w = log_2[pos[v] - pos[u] + 1];
    return mmin(ST[w][pos[u]], ST[w][pos[v] - (1 << w) + 1]);
}

inline int dist(int u, int v) {
    int Lca = lca(u, v);
    return dis[u] + dis[v] - dis[Lca] * 2;
}

inline void build() {
    for (int i = 1; i <= cnt; i ++)
        ST[0][i] = dfn[i];
    for (int i = 1; i < 20; i ++)
        for (int j = 1; j <= cnt; j ++)
            if (j + (1 << (i - 1)) > cnt) ST[i][j] = ST[i - 1][j];
            else ST[i][j] = mmin(ST[i - 1][j], ST[i - 1][j + (1 << (i - 1))]); 
}

int M;

struct node {
    int l, r, dis;
}tr[N << 1];

inline void update(int o, int o1, int o2) {
    static int d;
    static node tmp;
    if (tr[o1].dis == -1) {tr[o] = tr[o2]; return;}
    if (tr[o2].dis == -1) {tr[o] = tr[o1]; return;}
    if (tr[o1].dis > tr[o2].dis) tmp = tr[o1];
    else tmp = tr[o2]; 
    d = dist(tr[o1].l, tr[o2].l);
    if (d > tmp.dis) tmp.l = tr[o1].l, tmp.r = tr[o2].l, tmp.dis = d;
    d = dist(tr[o1].l, tr[o2].r);
    if (d > tmp.dis) tmp.l = tr[o1].l, tmp.r = tr[o2].r, tmp.dis = d;
    d = dist(tr[o1].r, tr[o2].l);
    if (d > tmp.dis) tmp.l = tr[o1].r, tmp.r = tr[o2].l, tmp.dis = d;
    d = dist(tr[o1].r, tr[o2].r);
    if (d > tmp.dis) tmp.l = tr[o1].r, tmp.r = tr[o2].r, tmp.dis = d;
    tr[o] = tmp;
}

inline void ask(int s, int t) {
    if (s > t) return;
    for (s += M - 1, t += M + 1; s ^ t ^ 1; s >>= 1, t >>= 1) {
        if (~s&1) update(0, 0, s ^ 1);
        if ( t&1) update(0, 0, t ^ 1);
    }
}

inline int get_char() {
    static const int SIZE = 1 << 23;
    static char *T, *S = T, buf[SIZE];
    if (S == T) {
        T = fread(buf, 1, SIZE, stdin) + (S = buf);
        if (S == T) return -1;
    }
    return *S ++;
}
 
inline void in(int &x) {
    static int ch;
    while (ch = get_char(), ch > 57 || ch < 48);x = ch - 48;
    while (ch = get_char(), ch > 47 && ch < 58) x = x * 10 + ch - 48;
}

int main() {
    int u, v, w, ans;
    log_2[1] = 0;
    for (int i = 2; i <= 200000; i ++) 
        if (i == 1 << (log_2[i - 1] + 1))
            log_2[i] = log_2[i - 1] + 1;
        else log_2[i] = log_2[i - 1];
    for (in(T); T --; ) {
        in(n), in(m), cnt = len = 0;
        for (int i = 1; i <= n; i ++)
            head[i] = 0;
        for (int i = 1; i < n; i ++) {
            in(ee[i].u), in(ee[i].v), in(ee[i].w);
            add(ee[i].u, ee[i].v, ee[i].w);
        }
        dfs1(1);
        for (M = 1; M < n + 2; M <<= 1);
        for (int i = 1; i <= n; i ++)
            tr[i + M].l = tr[i + M].r = dfn[i], tr[i + M].dis = 0;
        for (int i = n + M + 1; i <= (M << 1) + 1; i ++)
            tr[i].dis = -1;
        cnt = 0, dfs2(1), build();
        for (int i = M; i; i --) 
            update(i, i << 1, i << 1 | 1);
        for (int i = 1; i < n; i ++) 
            if (dep[ee[i].u] > dep[ee[i].v])
                swap(ee[i].u, ee[i].v);
        for (int u, v, i = 1; i <= m; i ++) {
            in(u), in(v), ans = 0;
            u = ee[u].v, v = ee[v].v, w = lca(u, v);
            if (w == u || w == v) {
                if (w != u) swap(u, v);
                tr[0].dis = -1, ask(1, st[u] - 1), ask(en[u] + 1, n), ans = max(ans, tr[0].dis);
                tr[0].dis = -1, ask(st[u], st[v] - 1), ask(en[v] + 1, en[u]), ans = max(ans, tr[0].dis);
                tr[0].dis = -1, ask(st[v], en[v]), ans = max(ans, tr[0].dis);
            }
            else {
                if (st[u] > st[v]) swap(u, v);
                tr[0].dis = -1, ask(1, st[u] - 1), ask(en[u] + 1, st[v] - 1), ask(en[v] + 1, n), ans = max(ans, tr[0].dis);
                tr[0].dis = -1, ask(st[u], en[u]), ans = max(ans, tr[0].dis);
                tr[0].dis = -1, ask(st[v], en[v]), ans = max(ans, tr[0].dis);
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}