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;

}