HDU 5877 Weak Pair(dfs + 树状数组 + 离散化)-优快云博客

本文链接：https://blog.youkuaiyun.com/hnust_Derker/article/details/52516206

本文介绍了一种使用树状数组求解特定值对数的算法。该算法适用于解决给定一棵树，求所有满足条件 a[u]*a[v] ≤ k 的节点对 (u, v) 的数量的问题。通过深度优先搜索结合树状数组维护每个深度的值来实现高效求解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<string>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#include<iostream>
typedef long long ll;
const int maxn = 1e5 + 10;
const double eps = 1e-7;
const ll INF = 1e18;
using namespace std;

ll C[maxn * 2], a[maxn * 2], b[maxn * 2];
vector<int> G[maxn];
int n, T, num, nn;
ll k;
int ch[maxn];
ll res;

void init() {
    num = 0; res = 0;
    for(int i = 0; i < maxn; i++)
        G[i].clear();
    memset(C, 0, sizeof(C));
    memset(ch, 0, sizeof(ch));
}

ll sum(int ind) {
    ll s = 0;
    while(ind) {
        s += C[ind]; ind -= ind & -ind;
    }
    return s;
}

void update(int ind, int add) {
    while(ind <= nn) {
        C[ind] += add; ind += ind & -ind;
    }
}

void dfs(int u, int d) {
    if(!a[u]) res += d;
    else {
        int ind = lower_bound(b, b + nn, k / a[u]) - b + 1;
        res += sum(ind);
    }
    int c = lower_bound(b, b + nn, a[u]) - b + 1;
    update(c, 1);
    for(int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        dfs(v, d + 1);
    }
    update(c, -1);
}


int main()
{
    scanf("%d", &T);
    while(T--) {
        init();
        scanf("%d %lld", &n, &k);
        for(int i = 1; i <= n; i++) {
            scanf("%lld", &a[i]);
            b[num++] = a[i];
            if(a[i]) b[num++] = k / a[i];
        }
        sort(b, b + num);
        nn = unique(b, b + num) - b;
        for(int i = 1; i < n; i++) {
            int f, t;
            scanf("%d %d", &f, &t);
            G[f].push_back(t);
            ch[t] = 1;
        }
        int node;
        for(int i = 1; i <= n; i++) {
            if(!ch[i]) { node = i; break; }
        }
        dfs(node, 0);
        printf("%lld\n", res);
    }
    return 0;
}

题意：给出一棵树，求出a[u] * a[v] <= k的（u，v）的对数

思路：搜索过程中用树状数组更新同深度的值，一个节点只能从根节点由一条路而来，相当于在一条路径上用树状数组