HDU5877

HDU5877 - Weak Pair

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做法：dfs的时候，用树状数组维护当前节点到跟节点的权值树状数组，离散化一下即可，类似统计树上逆序对。此题数据范围好像是假的，节点数开到200000可过。

#include <bits/stdc++.h>
#define pb push_back
typedef long long ll;
const int N = 2e5 + 7;
const ll inf = LONG_MAX;
using namespace std;
struct edge{int e,nxt;}E[N];
int h[N], cc;
void adde(int u,int v) {
    E[cc].e=v;E[cc].nxt=h[u];h[u]=cc;++cc;
}
int n, rt, in[N];
ll k, a[N];
vector<ll> v;
int num;
ll B[N], num0 , ans = 0;
int id(ll x) {
    return lower_bound(v.begin(),v.end(),x) - v.begin() + 1;
}
void add(int x,ll v) {
    for(int i=x;i<=num;i+=(i&(-i))) B[i]+=v;
}
ll ask(int x) {
    ll ans = 0;
    for(int i=x;i;i-=(i&(-i))) ans+=B[i];
    return ans;
}
void dfs(int u) {
    for(int i=h[u];~i;i=E[i].nxt) {
        int v = E[i].e;
        if(a[v]) ans += ask(id(k/a[v]));
        else ans += ask(id(inf));
        add(id(a[v]),1LL);
        dfs(v);
        add(id(a[v]),-1LL);
    }
}
void dfs2(int u,ll dep,ll num0) {
    for(int i=h[u];~i;i=E[i].nxt) {
        int v = E[i].e;
        if(a[v] == 0) ans += dep;
        else ans += num0;
        dfs2(v,dep+1,num0+(a[v]==0));
    }
}
int T;
int main() {
    scanf("%d",&T);
    while(T--) {
        scanf("%d%lld",&n,&k);
        v.clear();cc = 0;
        for(int i = 1; i <= n; ++i) {
            scanf("%lld",&a[i]);
            if(a[i]) v.pb(a[i]), v.pb(k/a[i]);
            else v.pb(0), v.pb(inf);
            h[i] = -1; in[i] = 0;
        }
        sort(v.begin(),v.end()); v.erase(unique(v.begin(),v.end()),v.end());
        num = v.size();
        for(int i = 1; i < n; ++i) {int u, v;
            scanf("%d%d", &u, &v);
            adde(u,v); ++in[v];
        }
        for(int i = 1; i <= n; ++i) if(!in[i]) {rt = i; break;}
        ans = 0;

        if(k == 0) {
            dfs2(rt,1,(a[rt]==0));
        }
        else {
            add(id(a[rt]),1LL);
            dfs(rt);
            add(id(a[rt]),-1LL);
        }
        printf("%lld
",ans);
    }
}