HDU 4123 Bob’s Race 树的直径+单调队列

题意：

给定n个点的带边权树Q个询问。

以下n-1行给出树

以下Q行每行一个数字表示询问。

首先求出dp[N] ：dp[i]表示i点距离树上最远点的距离

询问u， 表示求出 dp 数组中最长的连续序列使得序列中最大值-最小值 <= u，输出这个序列的长度。

思路：

求dp数组就是求个树的直径然后dfs一下。

对于每一个询问，能够用一个单调队列维护一下。O(n)的回答。

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <queue>
#include <algorithm>
#include <cmath>
using namespace std;
template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if(c=getchar(),c==EOF) return 0;
	while(c!='-'&&(c<'0'||c>'9')) c=getchar();
	sgn=(c=='-')?-1:1;
	ret=(c=='-')?0:(c-'0');
	while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
	ret*=sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
    if (x <0) {
        putchar('-');
        x = -x;
    }
    if(x>9) pt(x/10);
    putchar(x%10+'0');
}
typedef long long ll;
const int N = 50010;
int n, Q;
struct Edge{
    int to, nex;   ll dis;
}edge[N<<1];
struct node {
    int v, id;
    node() {}
    node(int _id, int _v) {
        id = _id; v = _v;
    }
};
int head[N], edgenum;
void init(){for(int i = 1; i <= n; i++)head[i] = -1; edgenum = 0;}
void add(int u, int v, ll d){
    Edge E = {v, head[u], d};
    edge[edgenum] = E;
    head[u] = edgenum++;
}
ll dis[N], dp[N], len;
int Stack[N], top, pre[N], vis[N];
int BFS(int x){
    for(int i = 1; i <= n; i++)
        dis[i] = -1;
    dis[x] = 0; pre[x] = -1;
    int far = x;
    queue<int> q; q.push(x);
    while(!q.empty())
    {
        int u = q.front(); q.pop();
        for(int i = head[u]; ~i; i = edge[i].nex){
            int v = edge[i].to;
            if(dis[v] == -1)
            {
                dis[v] = dis[u] + edge[i].dis;
                pre[v] = u;
                if(dis[far] < dis[v])
                    far = v;
                q.push(v);
            }
        }
    }
    return far;
}
void dfs(int u){
    vis[u] = 1;
    for(int i = head[u]; ~i; i = edge[i].nex)
    {
        int v = edge[i].to;
        if(vis[v])continue;
        dp[v] = dp[u] + edge[i].dis;
        dfs(v);
    }
}
void build(){//预处理树的直径
    int E = BFS(1);
    int S = BFS(E);
    top = 0;
    int u = S;
    len = dis[S];
    for(int i = 1; i <= n; i++) vis[i] = 0;
    while(u!=-1)
    {
        Stack[top++] = u;
        dp[u] = max(dis[u], len - dis[u]);
        vis[u] = 1;
        u = pre[u];
    }
    for(int i = 0; i < top; i++)     dfs(Stack[i]);
}
void input(){
    init();  ll d;
    for(int i = 1, u, v; i < n; i++)
    {
        rd(u); rd(v); rd(d);
        add(u, v, d); add(v, u, d);
    }
}

node mx[N], mi[N];
int h1, t1, h2, t2;
int main() {
    int v, idx, ans;
	while(cin>>n>>Q, n+Q) {
		input();
		build();
		while(Q--)
        {
            rd(v);
            ans = h1 = t1 = h2 = t2 = 0;
            idx = 1;
            for (int i = 1; i <= n; ++i) {
                while (h1!=t1 && mx[t1-1].v <= dp[i])
                    -- t1;
                mx[t1++] = node(i, dp[i]);
                while (h2!=t2 && mi[t2-1].v >= dp[i])
                    -- t2;
                mi[t2++] = node(i, dp[i]);
                while (h1!=t1&&h2!=t2) {
                    if (mx[h1].v-mi[h2].v>v)
                        ++ idx;
                    else
                        break;
                    while (h1!=t1&&mx[h1].id<idx)
                        ++h1;
                    while (h2!=t2&&mi[h2].id<idx)
                        ++h2;
                }
                ans = max(ans, i-idx+1);
            }
            pt(ans);
            putchar('\n');
        }
	}
	return 0;
}