HDU 5877 Weak Pair(dfs + 树状数组 + 离散化)

#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