hdu 5877/ 2016 ACM/ICPC Dalian Online 1010 Weak Pair

题目链接

分析：树上的节点祖先与儿子的关系，一般就会想到dfs序。正解就是对树先进行dfs序排列，再将问题转化到树状数组统计个数。应该把节点按照权值从大到小排序，这样对于a[i]，K/a[i]就是从小到大的顺序。这样更新树状数组就不会造成计算的混乱了。

多组数据没有每次先清除边我真是太智障了。

代码：

/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define   offcin        ios::sync_with_stdio(false)
#define   sigma_size    26
#define   lson          l,m,v<<1
#define   rson          m+1,r,v<<1|1
#define   slch          v<<1
#define   srch          v<<1|1
#define   sgetmid       int m = (l+r)>>1
#define   LL            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   pb            push_back
#define   fi            first
#define   se            second
const int    INF    = 0x3f3f3f3f;
const LL     INFF   = 1e18;
const double pi     = acos(-1.0);
const double inf    = 1e18;
const double eps    = 1e-9;
const LL     mod    = 1e9+7;
const int    maxmat = 10;
const ull    BASE   = 31;
/*****************************************************/
const int maxn = 1e5 + 5;
int in[maxn], out[maxn], dfs_clock;
int a[maxn], outdgree[maxn];
int sum[maxn];
std::vector<int> G[maxn];
int N;
LL K;
struct Node {
    int val, id;
    bool operator <(const Node &rhs) const {
        return val < rhs.val;
    }
}node[maxn];
void init() {
    mem(sum, 0);
    mem(node, 0);
    mem(in, 0);
    mem(out, 0);
    mem(outdgree, 0);
    dfs_clock = 0;
}
void dfs(int u, int fa) {
    if (in[u]) return;
    in[u] = ++ dfs_clock;
    for (int i = 0; i < G[u].size(); i ++) {
        int v = G[u][i];
        if (v == fa) continue;
        dfs(v, u);
    }
    out[u] = dfs_clock;
}
void add(int x) {
    while (x <= N) {
        sum[x] ++;
        x += lowbit(x);
    }
}
int query(int x) {
    int res = 0;
    while (x) {
        res += sum[x];
        x -= lowbit(x);
    }
    return res;
}
int main(int argc, char const *argv[]) {
    int T;
    cin>>T;
    while (T --) {
        init();
        scanf("%d%I64d", &N, &K);
        for (int i = 1; i <= N; i ++) G[i].clear();
        for (int i = 1; i <= N; i ++) {
            scanf("%d", &node[i].val);
            node[i].id = i;
        }
        for (int i = 0; i < N - 1; i ++) {
            int u, v;
            scanf("%d%d", &u, &v);
            outdgree[v] ++;
            G[u].pb(v);
            G[v].pb(u);
        }
        sort(node + 1, node + N + 1);
        int root = 1;
        for (int i = 1; i <= N; i ++) if (!outdgree[i]) {
            root = i;
            break;
        }
        dfs(root, -1);
        int pos = 1;
        LL ans = 0;
        for (int i = N; i >= 1; i --) {
            LL tmp = K / (LL)node[i].val;
            int id = node[i].id;
            while (node[pos].val <= tmp && pos <= N)
                add(in[node[pos ++].id]);
            ans += query(out[id]) - query(in[id]);
        }
        cout<<ans<<endl;
    }
    return 0;
}