HUD6162(树链剖分)

最新推荐文章于 2025-03-04 08:46:39 发布

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本文介绍了一种利用离线处理结合树链剖分解决特定路径问题的方法，通过将问题转化为区间查询，实现对树上节点价值的有效计算。

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Ch’s gift

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 613 Accepted Submission(s): 212

Problem Description

Mr. Cui is working off-campus and he misses his girl friend very much. After a whole night tossing and turning, he decides to get to his girl friend's city and of course, with well-chosen gifts. He knows neither too low the price could a gift be since his girl friend won't like it, nor too high of it since he might consider not worth to do. So he will only buy gifts whose price is between [a,b].
There are n cities in the country and (n-1) bi-directional roads. Each city can be reached from any other city. In the ith city, there is a specialty of price ci Cui could buy as a gift. Cui buy at most 1 gift in a city. Cui starts his trip from city s and his girl friend is in city t. As mentioned above, Cui is so hurry that he will choose the quickest way to his girl friend(in other words, he won't pass a city twice) and of course, buy as many as gifts as possible. Now he wants to know, how much money does he need to prepare for all the gifts?

Input

There are multiple cases.

For each case:
The first line contains tow integers n,m(1≤n,m≤10^5), representing the number of cities and the number of situations.
The second line contains n integers c1,c2,...,cn(1≤ci≤10^9), indicating the price of city i's specialty.
Then n-1 lines follows. Each line has two integers x,y(1≤x,y≤n), meaning there is road between city x and city y.
Next m line follows. In each line there are four integers s,t,a,b(1≤s,t≤n;1≤a≤b≤10^9), which indicates start city, end city, lower bound of the price, upper bound of the price, respectively, as the exact meaning mentioned in the description above

Output

Output m space-separated integers in one line, and the ith number should be the answer to the ith situation.

Sample Input

Sample Output

  
   7 1 4

Source

2017 Multi-University Training Contest - Team 9

解题思路：比赛时知道要离线处理，也知道大概是树链剖分或者dfs序之类的东西，但是对于链上的每个点的权值要在[l, r]之间这里我不知道怎样处理，其实对于这个问题，我们只需要把这个区间分为两部分[1, l - 1], [1, r],然后二者相减就行，然后我们把所有点的权值从小到大排序，把所有询问按位置大小从小到大排个序，之后用树链剖分弄一下就行。。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 100000 + 10;
struct query{
    int id;
    LL flag;
    int s, t;
    LL sum;
    LL loc;
    query(){
        sum = 0;
    }
    bool operator <(const query &res) const{
        return loc < res.loc;
    }

}Query[maxn<<1];
struct node{
    LL value;
    int id;
    bool operator <(const node &res) const{
        return value < res.value;
    }
}Node[maxn];
int n, m;
LL re[maxn];
LL Tree[maxn];
int root;
vector<int> g[maxn];
int depth[maxn];
int father[maxn];
int Rank[maxn];
int son[maxn];
int Size[maxn];
int head[maxn];
int cnt;
int querysum;
int Hash[maxn][2];
int lowbit(int x)
{
    return x&(-x);
}
void update(int loc, LL value)
{
    for(int i = loc; i <= n; i += lowbit(i))
    {
        Tree[i] += value;
    }
}
LL get(int loc)
{
    LL ans = 0;
    for(int i = loc; i >= 1; i -= lowbit(i))
    {
        ans += Tree[i];
    }
    return ans;
}
void dfs1(int u, int d, int f)
{
    depth[u] = d;
    Size[u] = 1;
    son[u] = -1;
    int Max = -1;
    for(int i = 0; i < g[u].size(); i++)
    {
        int v = g[u][i];
        if(v == f) continue;
        dfs1(v, d + 1, u);
        father[v] = u;
        if(Size[v] > Max)
        {
            Max = Size[v];
            son[u] = v;
        }
        Size[u] += Size[v];

    }
}
void dfs2(int u, int f, int fp)
{
    head[u] = fp;
    Rank[u] = ++cnt;
    if(son[u] != -1)
    {
        dfs2(son[u], u, fp);
    }
    for(int i = 0; i < g[u].size(); i++)
    {
        int v = g[u][i];
        if(v == f || v == son[u]) continue;
        dfs2(v, u, v);
    }
}
LL querySum(int l, int r)
{
    return get(Rank[r]) - get(Rank[l] - 1);
}
LL solveSum(int u, int v)
{
    int f1 = head[u];
    int f2 = head[v];
    LL ans = 0;
    while(f1 != f2)
    {
        if(depth[f1] > depth[f2])
        {
            ans += querySum(f1, u);
            u = father[f1];
        }
        else
        {
            ans += querySum(f2, v);
            v = father[f2];
        }
        f1 = head[u];
        f2 = head[v];
    }
    if(depth[u] > depth[v])
    {
       ans += querySum(v, u);
    }
    else ans += querySum(u, v);
    return ans;
}
void init()
{
    memset(Tree, 0, sizeof(Tree));
    memset(re, 0, sizeof(re));
    root = 1;
    cnt = 0;
    querysum = 0;
    for(int i = 1; i <= n; i++)
    {
        g[i].clear();
    }
}
int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        init();
        for(int i = 1; i <= n; i++)
        {
            scanf("%lld", &Node[i].value);
            Node[i].id = i;
        }
        sort(Node + 1, Node + n + 1);
        int x, y;
        for(int i = 1; i < n; i++)
        {
            scanf("%d%d", &x, &y);
            g[x].push_back(y);
            g[y].push_back(x);
        }
        dfs1(root, 0, root);
        dfs2(root, root, root);
        int s, t;
        LL a, b;
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d%lld%lld", &s, &t, &a, &b);
            Query[++querysum].loc = a - 1;
            Query[querysum].id = i;
            Query[querysum].s = s;
            Query[querysum].t = t;
            Query[querysum].sum = 0;
            Query[querysum].flag = -1;
            Query[++querysum].loc = b;
            Query[querysum].id = i;
            Query[querysum].s = s;
            Query[querysum].t = t;
            Query[querysum].sum = 0;
            Query[querysum].flag = 1;
        }
        sort(Query + 1, Query + querysum + 1);
        int p1 = 1;
        int p2 = 1;
        while(p2 <= n && p1 <= querysum)
        {
            while(Node[p2].value <= Query[p1].loc && p2 <= n)
            {
                update(Rank[Node[p2].id], Node[p2].value);
                p2++;
            }
            Query[p1].sum = solveSum(Query[p1].s, Query[p1].t);
            p1++;
        }
        while(p1 <= querysum)
        {
            Query[p1].sum = solveSum(Query[p1].s, Query[p1].t);
            p1++;
        }
        for(int i = 1; i <= querysum; i++)
        {
            re[Query[i].id] += Query[i].sum * Query[i].flag;
        }
        for(int i = 1; i <= m; i++)
        {
            if(i == m) printf("%lld\n", re[i]);
            else printf("%lld ", re[i]);
        }
    }
    return 0;
}