HDU5877Weak Pair

                                                Weak Pair

                                Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
                                         

Problem Description

You are given a rooted tree of N nodes, labeled from 1 to N. To the ith node a non-negative value ai is assigned.An ordered pair of nodes (u,v) is said to be weak if
  (1) u is an ancestor of v (Note: In this problem a node u is not considered an ancestor of itself);
  (2) au×av≤k.

Can you find the number of weak pairs in the tree?

 

Input

There are multiple cases in the data set.
  The first line of input contains an integer T denoting number of test cases.
  For each case, the first line contains two space-separated integers, N and k, respectively.
  The second line contains N space-separated integers, denoting a1 to aN.
  Each of the subsequent lines contains two space-separated integers defining an edge connecting nodes u and v , where node u is the parent of node v.

  Constrains:
  
  1≤N≤105
  
  0≤ai≤109
  
  0≤k≤1018

 

Output

For each test case, print a single integer on a single line denoting the number of weak pairs in the tree.

 

Sample Input

1
2 3
1 2
1 2

 

Sample Output

1

 

Source

2016 ACM/ICPC Asia Regional Dalian Online

题意十分明确， 就是求出符合题意的有序点对个数。

首先对ai离散,离散之后的结果用rk[i]表示，然后进行二分预处理得到f[i],其中f[i]的意义为:其他的点和i这个节点满足weakpair要求的权值最大名次(名次权值小的排在前面)。

然后就开始跑一遍DFS，树状数组维护一下答案，就好了。

#include <bits/stdc++.h>

using namespace std;

#define REP(i,n)                for(int i(0); i <  (n); ++i)
#define rep(i,a,b)              for(int i(a); i <= (b); ++i)
#define dec(i,a,b)              for(int i(a); i >= (b); --i)
#define for_edge(i,x)           for(int i = H[x]; i; i = X[i])

#define LL      long long
#define ULL     unsigned long long
#define MP      make_pair
#define PB      push_back
#define FI      first
#define SE      second
#define INF     1 << 30

const int N     =    300000      +       10;
const int M     =    10000       +       10;
const int Q     =    1000        +       10;
const int A     =    30          +       1;


struct Node{
	LL num;
	int id;
	friend bool operator < (const Node &a, const Node &b){
		return a.num < b.num;
	}
} tree[N];

int E[N << 1], X[N << 1], H[N << 1];
LL a[N];
int T, et;
int n;
LL k;
int x, y;
LL rk[N], f[N];
LL now;
int l, r;
bool pa[N];
int root;
LL c[N];
LL ans;
bool v[N];

inline void addedge(int a, int b){
	E[++et] = b, X[et] = H[a], H[a] = et;
}

inline void add(LL x, LL val){ 
	for (; x <= n; x += (x) & (-x)) 
		c[x] += val;
}

inline LL query(LL x){
	LL ret(0); 
	for (; x; x -= (x) & (-x)) ret += c[x]; 
	return ret;
}

void dfs(int x){
	add(rk[x], 1);
	for_edge(i, x) if (!v[E[i]]) dfs(E[i]), v[E[i]] = true;
	add(rk[x], -1);
	ans += query(f[x]);
}


int main(){

	scanf("%d", &T);
	while (T--){
		et = 0;
		scanf("%d%lld", &n, &k);
		rep(i, 1, n) scanf("%lld", a + i);
		memset(v, false, sizeof v);
		memset(pa, true, sizeof pa);
		memset(tree, 0, sizeof tree);
		memset(H, 0, sizeof H);
		rep(i, 1, n - 1){
			scanf("%d%d", &x, &y);
			addedge(x, y);
			pa[y] = false;
		}

		rep(i, 1, n){
			tree[i].num = a[i];
			tree[i].id = i;
		}

		sort(tree + 1, tree + n + 1);
		rk[tree[1].id] = 1;
		rep(i, 2, n) 
			if (tree[i].num == tree[i - 1].num) rk[tree[i].id] = rk[tree[i - 1].id];
			else rk[tree[i].id] = rk[tree[i - 1].id] + 1;

		rep(i, 1, n){
			now = k / a[i];
			l = 1; r = n;
			if (tree[1].num > now) f[i] = 0; 
			else{	
				while (l + 1 < r){
					int mid = (l + r) >> 1;
					if (tree[mid].num <= now) l = mid;
					else r = mid - 1;
				}

				if (tree[r].num <= now) l = r;
				f[i] = rk[tree[l].id];
			}

		}
		root = 0;
		rep(i, 1, n) if (pa[i]){ root = i; break;}
		memset(c, 0, sizeof c);
		ans = 0;
		v[root] = true;
		dfs(root);
		printf("%lld\n", ans);


	}



	return 0;

}