Apple Tree(树状搜索,树形DP)

合集 - 刷题集(97)
1.妖梦拼木棒2.[USACO1.2]回文平方数 Palindromic Squares3.[COCI2012-2013#1] LJUBOMORA4.火柴棒等式5.Kieszonkowe6.[COCI2014-2015#5] TRAKTOR7.硬币翻转8.[蓝桥杯 2020 省 AB1] 走方格9.P2097 资料分发10.[AHOI2005]约数研究11.[NOIP2004 普及组] 火星人12.我焯,原神(这真是个题)13.滑雪 OJ365114.CV的人有福啦!YTU贪心训练2(部分注释)15.砝码称重16.警钟长鸣-----找零钱17.[USACO09OCT]Barn Echoes G18.【CSGRound2】光骓者的荣耀19.初见蓝桥杯20.[蓝桥杯 2022 省 B] 扫雷21.开餐馆22.P9241 [蓝桥杯 2023 省 B] 飞机降落23.P7083 [NWRRC2013] Energy Tycoon24.高精度25.P7954 [COCI2014-2015#6] PAPRIKA26.(输出路径搜索)[USACO06OCT] Cows on Skates G27.畜栏保留问题28.(贪心+搜索+剪枝)P8801 [蓝桥杯 2022 国 B] 最大数字29.蓝桥杯 入门----普及-30.[NOIP2016 提高组] 玩具谜题31.[NOIP2002 提高组] 字串变换(无向图查重)32.[NOIP2000 提高组] 单词接龙33.POJ 硬币34.动态规划学习 1(最长上升子序列问题)35.动态规划学习 2(背包问题及其变形)36. 切割回文37.[NOIP2001 提高组] 数的划分(剪枝)38.字符串转换数字,sscanf和sprintf大法39.[CTSC1997] 选课(树状DP)40. k 倍区间(同余定理,组合数)41.[蓝桥杯 2018 省 AB] 全球变暖42.马的遍历(对bfs的更深一层探讨)43.找零钱44.Pasha Maximizes45.Newspaper Headline46.Balanced Ternary String47.(数论)素数,质数48.(数论) 约数49.Helping the Nature50.(博弈论)Even Number Addicts51.Present52.Mr. Kitayuta's Colorful Graph(二维并查集,弱化版)53.Edgy Trees(dfs,并查集,快速幂,树形结构,红黑树)54.并查集55.再谈 前缀和,差分,离散化56.Best Cow Fences(前缀和+特殊二分)57.二分58.同类型,类背包动态规划,选地dp59.Primes on Interval(欧拉筛+二分+滑动窗口)60.字符串(KMP , Trie树 , STL)61.Codeforces1 #879 div.262.Add Modulo 10(数论,思维,数学,规律)63.期末考试YTU4035: Shmily(数学,等差数列)64.[POI2006] OKR-Periods of Words65.Getting Zero(Bfs)66.快速幂67.Dreaming of Freedom(数论,贪心)68.Scoring Subsequences69.Dolce Vita70.Cake Assembly Line71.Constructive Problem72.Forever Winter73.Maximum Sum(前缀和)74.Contrast Value(数学规律)75.Lamps(STL+双端队列)76.Long Long77.Sum in Binary Tree
78.Apple Tree(树状搜索,树形DP)
79.Strong Password(贪心思想)80.Come Together81.Notepad82.Hamiltonian Wall83.7.17作业84.逛画展(双指针)85.#88586.树状数组87.Subsequence Addition88.7.22作业89.Diverse Substrings90.Tracking Segments(二分,区间前缀)91.最大食物链计数92.线段树193.[TJOI2007] 线段(状态机)94.#87895.测试196.李白打酒97.砍竹子
收起
Apple Tree
time limit per test
4 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output

Timofey has an apple tree growing in his garden; it is a rooted tree of n vertices with the root in vertex 11 (the vertices are numbered from 11 to n). A tree is a connected graph without loops and multiple edges.

This tree is very unusual — it grows with its root upwards. However, it's quite normal for programmer's trees.

The apple tree is quite young, so only two apples will grow on it. Apples will grow in certain vertices (these vertices may be the same). After the apples grow, Timofey starts shaking the apple tree until the apples fall. Each time Timofey shakes the apple tree, the following happens to each of the apples:

Let the apple now be at vertex u.

  • If a vertex u has a child, the apple moves to it (if there are several such vertices, the apple can move to any of them).
  • Otherwise, the apple falls from the tree.

It can be shown that after a finite time, both apples will fall from the tree.

Timofey has q assumptions in which vertices apples can grow. He assumes that apples can grow in vertices x and y, and wants to know the number of pairs of vertices (a, b) from which apples can fall from the tree, where a — the vertex from which an apple from vertex x will fall, b — the vertex from which an apple from vertex y will fall. Help him do this.

Input

The first line contains integer t (1≤t≤1041≤≤104) — the number of test cases.

The first line of each test case contains integer n (2≤n≤2⋅1052≤≤2⋅105) — the number of vertices in the tree.

Then there are n−1−1 lines describing the tree. In line i there are two integers ui and vi (1≤ui,vi≤n1≤,≤, ui≠vi≠) — edge in tree.

The next line contains a single integer q (1≤q≤2⋅1051≤≤2⋅105) — the number of Timofey's assumptions.

Each of the next q lines contains two integers xi and yi (1≤xi,yi≤n1≤,≤) — the supposed vertices on which the apples will grow for the assumption i.

It is guaranteed that the sum of n does not exceed 2⋅1052⋅105. Similarly, It is guaranteed that the sum of q does not exceed 2⋅1052⋅105.

Output

For each Timofey's assumption output the number of ordered pairs of vertices from which apples can fall from the tree if the assumption is true on a separate line.

Examples
input
Copy
2
5
1 2
3 4
5 3
3 2
4
3 4
5 1
4 4
1 3
3
1 2
1 3
3
1 1
2 3
3 1
output
Copy
2
2
1
4
4
1
2
input
Copy
2
5
5 1
1 2
2 3
4 3
2
5 5
5 1
5
3 2
5 3
2 1
4 2
3
4 3
2 1
4 2
output
Copy
1
2
1
4
2
Note

In the first example:

  • For the first assumption, there are two possible pairs of vertices from which apples can fall from the tree: (4,4),(5,4)(4,4),(5,4).
  • For the second assumption there are also two pairs: (5,4),(5,5)(5,4),(5,5).
  • For the third assumption there is only one pair: (4,4)(4,4).
  • For the fourth assumption, there are 44 pairs: (4,4),(4,5),(5,4),(5,5)(4,4),(4,5),(5,4),(5,5).
Tree from the first example.

For the second example, there are 44 of possible pairs of vertices from which apples can fall: (2,3),(2,2),(3,2),(3,3)(2,3),(2,2),(3,2),(3,3). For the second assumption, there is only one possible pair: (2,3)(2,3). For the third assumption, there are two pairs: (3,2),(3,3)(3,2),(3,3).

复制代码
// Apple Tree
// 树形搜索,建立根节点和子节点,dfs内u代表当前的节点,v代表父节点
// 如果搜到父节点的话就返回
// 使用邻接表建立树
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e6+10,mod=1e9+7;
string s;
int n,t,a[N],f[N],res,num,ans,m,q;
int idx,h[N],e[N],ne[N];
bool vis[N];
void add(int a,int b)
{
    e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}
void dfs(int u,int v)
{
    bool flag=false;
    for(int i=h[u];i!=-1;i=ne[i]){
        int x=e[i];
        if(e[i]==v) continue;
        flag=true,dfs(x,u);
        f[u]+=f[x];
    }
    if(!flag) f[u]=1;
}
signed main()
{
    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        cin>>n;
        for(int i=0;i<=n+5;i++) h[i]=-1,e[i]=ne[i]=f[i]=0;
        for(int i=1;i<n;i++){
            int x,y;
            cin>>x>>y;
            add(x,y),add(y,x);
        }
        dfs(1,0);
        cin>>m;
        while(m--){
            cin>>ans>>num;
            cout<<f[ans]*f[num]<<endl;
        }
    }
    return 0;
}
复制代码

 


__EOF__

本文作者：Sakurajimamai
本文链接：https://www.cnblogs.com/o-Sakurajimamai-o/p/17544897.html
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