标签：
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 |
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++】
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <map> 5 #define LL __int64 6 #define mx 100005 7 std::map<LL, int> mp; 8 LL data[mx], dMP[mx], opt, k; 9 int n; 10 11 struct Edge{ 12 int to, next; 13 }edge[mx]; 14 int iE, head[mx], d[mx], tr[mx << 2], fid; 15 void addEdge(int u, int v){ 16 edge[iE].to = v; edge[iE].next = head[u]; head[u] = iE++; 17 } 18 19 void cnt(int l, int r, int now){ 20 if (r <= fid){ opt += tr[now]; return; } 21 int mid = l + r >> 1; 22 if (fid <= mid) return cnt(l, mid, now << 1); 23 opt += tr[now << 1]; return cnt(mid + 1, r, now << 1 | 1); 24 } 25 void sub(int l, int r, int now){ 26 --tr[now]; 27 if (l == r) return; 28 int mid = l + r >> 1; 29 if (fid <= mid) return sub(l, mid, now << 1); 30 return sub(mid + 1, r, now << 1 | 1); 31 } 32 void add(int l, int r, int now){ 33 ++tr[now]; 34 if (l == r) return; 35 int mid = l + r >> 1; 36 if (fid <= mid) return add(l, mid, now << 1); 37 return add(mid + 1, r, now << 1 | 1); 38 } 39 40 void DFS(int now){ 41 int u; 42 LL temp; 43 if (data[now] == 0 || (temp = k / data[now]) >= dMP[n]) fid = n; 44 else fid = std::upper_bound(dMP + 1, dMP + 1 + n, temp) - dMP - 1; 45 if (fid) cnt(1, n, 1); 46 fid = mp[data[now]]; add(1, n, 1); 47 for (u = head[now]; ~u; u = edge[u].next) DFS(edge[u].to); 48 fid = mp[data[now]]; sub(1, n, 1); 49 } 50 51 int main(){ 52 int t, i, u, v; 53 scanf("%d", &t); 54 while (t--){ 55 mp.clear(); memset(tr, 0, sizeof(tr)); 56 scanf("%d%I64d", &n, &k); 57 for (i = 1; i <= n; ++i) scanf("%I64d", &data[i]); 58 memcpy(dMP, data, sizeof(data)); 59 std::sort(dMP + 1, dMP + n + 1); 60 for (i = 1; i <= n; ++i) mp[dMP[i]] = i; 61 62 memset(head, -1, sizeof(head)); memset(d, 0, sizeof(d)); 63 iE = 0; 64 for (i = 1; i < n; ++i){ 65 scanf("%d%d", &u, &v); 66 addEdge(u, v); ++d[v]; 67 } 68 69 for (i = 1; d[i]; ++i); 70 opt = 0; DFS(i); 71 printf("%I64d\n", opt); 72 } 73 return 0; 74 }
标签：
原文地址：http://www.cnblogs.com/Simon-X/p/5883563.html
