HDU 5877 Weak Pair（弱点对）

HDU 5877 Weak Pair（弱点对）

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

Description	题目描述
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?	给你有N个节点的有根树，编号从1到N。第i个节点会被分配一个非负数ai。一个满足如下条件的有序点对(u, v)则被认为是弱点对 (1)u是v的先祖节点(注意：这个问题中u不能为自身的先祖节点)。 (2) au X av ≤ k 你能找出这棵树里有多少弱点对吗？

Input

输入

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

多组测试用例。 输入的第一行有一个整数T表示测试用例的数量。 对于每个测试用例，第一行有两个用空格分隔的整数，N与k， 第二行有N个用空格分隔的整数，表示a1到aN。 随后每(N-1)行有两个用空格分隔的数u与v表示连接两节点的一条边，其中u是v的父节点。 范围如下: 1≤N≤105 0≤ai≤109 0≤k≤1018

Output	输出
For each test case, print a single integer on a single line denoting the number of weak pairs in the tree.	对于每个测试用例，输出一个表示树中弱点对数量的整数在单独一行。

Sample Input - 输入样例	Sample Output - 输出样例
1 2 3 1 2 1 2	1

【题解】

DFS + 离散化线段树

用DFS从上往下（从根往叶子）添加与释放节点进线段树，利用线段树查找符合的区间中元素数量。

因为单个数的值比较大，但是总数比较少，线段树还算可以接受。根据总数进行离散化，其实就是排个序+map。出于想用upper_bound的强迫症没有把k/ai的结果一并加入离散化（注意0），不然代码量应该更少。

【代码 C++】

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <map>
 #define LL __int64
 #define mx 100005
 std::map<LL, int> mp;
 LL data[mx], dMP[mx], opt, k;
 int n;

 struct Edge{
     int to, next;
 }edge[mx];
 int iE, head[mx], d[mx], tr[mx << ], fid;
 void addEdge(int u, int v){
     edge[iE].to = v; edge[iE].next = head[u]; head[u] = iE++;
 }

 void cnt(int l, int r, int now){
     if (r <= fid){ opt += tr[now]; return; }
     int mid = l + r >> ;
     if (fid <= mid) return cnt(l, mid, now << );
     opt += tr[now << ]; return cnt(mid + , r, now <<  | );
 }
 void sub(int l, int r, int now){
     --tr[now];
     if (l == r) return;
     int mid = l + r >> ;
     if (fid <= mid) return sub(l, mid, now << );
     return sub(mid + , r, now <<  | );
 }
 void add(int l, int r, int now){
     ++tr[now];
     if (l == r) return;
     int mid = l + r >> ;
     if (fid <= mid) return add(l, mid, now << );
     return add(mid + , r, now <<  | );
 }

 void DFS(int now){
     int u;
     LL temp;
     if (data[now] ==  || (temp = k / data[now]) >= dMP[n]) fid = n;
     else fid = std::upper_bound(dMP + , dMP +  + n, temp) - dMP - ;
     if (fid) cnt(, n, );
     fid = mp[data[now]]; add(, n, );
     for (u = head[now]; ~u; u = edge[u].next) DFS(edge[u].to);
     fid = mp[data[now]]; sub(, n, );
 }

 int main(){
     int t, i, u, v;
     scanf("%d", &t);
     while (t--){
         mp.clear(); memset(tr, , sizeof(tr));
         scanf("%d%I64d", &n, &k);
         for (i = ; i <= n; ++i) scanf("%I64d", &data[i]);
         memcpy(dMP, data, sizeof(data));
         std::sort(dMP + , dMP + n + );
         for (i = ; i <= n; ++i) mp[dMP[i]] = i;

         memset(head, -, sizeof(head)); memset(d, , sizeof(d));
         iE = ;
         for (i = ; i < n; ++i){
             scanf("%d%d", &u, &v);
             addEdge(u, v); ++d[v];
         }

         for (i = ; d[i]; ++i);
         opt = ; DFS(i);
         printf("%I64d\n", opt);
     }
     return ;
 }