学步园

题目链接：hdu 4123 Bob’s Race

题目大意：一个城镇有N个住户，N-1条路连接两个住户，保证N个住户联通，M次询问，给定N条边的信息，包括连

接的住户序号以及路的长度。然后是M次询问，每次询问Q，要求找到最长的连续序号，使得Max(dis[i]) - Min(dis[i]) ≤ 

Q(l≤i≤r)，输出最大的r-l+1。dis[i]为从第i个住户出发，不重复走过路能移动的最远距离。

解题思路：树形dp，通过两次dfs，第1次处理出每个节点中孩子节点移动的最长距离和第二长距离，第2次考虑从父

亲节点过来的路径，维护每个节点的最长距离即可。

然后用RMQ与处理加快询问速度，枚举右边界确定左边界，单次复杂度为o(n).

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 50005;

int N, M, E, Q, first[maxn], jump[maxn * 2], dpMax[maxn][20], dpMin[maxn][20];
struct Edge {
    int u, v, w;
    void set(int u, int v, int w) {
        this->u = u;
        this->v = v;
        this->w = w;
    }
}ed[maxn * 2];

inline void add_Edge (int u, int v, int w) {
    ed[E].set(u, v, w);
    jump[E] = first[u];
    first[u] = E++;
}

int fmax[maxn], fidx[maxn], smax[maxn], sidx[maxn];

inline void maintain(int u, int w, int v) {
    if (w > smax[u]) {
        smax[u] = w;
        sidx[u] = v;
    }

    if (smax[u] > fmax[u]) {
        swap(smax[u], fmax[u]);
        swap(sidx[u], fidx[u]);
    }
}

void dfs(int u, int pre) {
    fmax[u] = fidx[u] = smax[u] = sidx[u] = 0;
    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = ed[i].v;
        if (v == pre)
            continue;
        dfs(v, u);
        maintain(u, fmax[v] + ed[i].w, v);
    }
}

void dfs(int u, int pre, int d) {
    maintain(u, d, pre);
    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = ed[i].v;
        if (v == pre)
            continue;
        dfs(v, u, (v == fidx[u] ? smax[u] : fmax[u]) + ed[i].w);
    }
}

void rmq_init() {
    for (int i = 1; i <= N; i++)
        dpMax[i][0] = dpMin[i][0] = fmax[i];

    for (int k = 1; (1<<k) <= N; k++) {
        for (int i = 1; i + (1<<k) - 1 <= N; i++) {
            dpMax[i][k] = max(dpMax[i][k-1], dpMax[i+(1<<(k-1))][k-1]);
            dpMin[i][k] = min(dpMin[i][k-1], dpMin[i+(1<<(k-1))][k-1]);
        }
    }
}

int rmq_query(int l, int r) {
    int k = 0;
    while ((1<<(k+1)) <= r - l + 1) k++;
    return max(dpMax[l][k], dpMax[r-(1<<k)+1][k]) - min(dpMin[l][k], dpMin[r-(1<<k)+1][k]);
}

int main () {
    while (scanf("%d%d", &N, &M) == 2 && N + M) {
        int u, v, w;
        E = 0;
        memset(first, -1, sizeof(first));
        for (int i = 1; i < N; i++) {
            scanf("%d%d%d", &u, &v, &w);
            add_Edge(u, v, w);
            add_Edge(v, u, w);
        }

        dfs(1, 0);
        dfs(1, 0, 0);

        rmq_init();
        while (M--) {
            int ans = 0, mv = 1;
            scanf("%d", &Q);
            for (int i = 1; i <= N; i++) {
                while (mv <= i && rmq_query(mv , i) > Q) mv++;
                ans = max(ans, i - mv + 1);
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}