HDOJ 5416 CRB and Tree DFS

CRB and Tree

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 690    Accepted Submission(s): 221

Problem Description

CRB has a tree, whose vertices are labeled by 1, 2, …, N. They are connected by N – 1 edges. Each edge has a weight.
For any two vertices u and v(possibly equal), f(u,v) is xor(exclusive-or) sum of weights of all edges on the path from u to v.
CRB’s task is for given s, to calculate the number of unordered pairs (u,v) such that f(u,v) = s. Can you help him?

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains an integer N denoting the number of vertices.
Each of the next N - 1 lines contains three space separated integers a, b and c denoting an edge between a and b, whose weight is c.
The next line contains an integer Q denoting the number of queries.
Each of the next Q lines contains a single integer s.
1 ≤ T ≤ 25
1 ≤ N ≤ 105
1 ≤ Q ≤ 10
1 ≤ a, b ≤ N
0 ≤ c, s ≤ 105
It is guaranteed that given edges form a tree.

 

Output

For each query, output one line containing the answer.

 

Sample Input

1
3
1 2 1
2 3 2
3
2
3
4

 

Sample Output

1
1
0
Hint
For the first query, (2, 3) is the only pair that f(u, v) = 2.
For the second query, (1, 3) is the only one.
For the third query, there are no pair (u, v) such that f(u, v) = 4.
 

 

Author

KUT（DPRK）

 

Source

2015 Multi-University Training Contest 10

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月21日 星期五 14时10分39秒
File Name     :HDOJ5416.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;
const int maxn=100100;
const int MX=1e6+10;

int n,Q;

struct Edge
{
	int to,next,val;
}edge[maxn*2];

int Adj[maxn],Size;

void init()
{
	memset(Adj,-1,sizeof(Adj)); Size=0;
}

void Add_Edge(int u,int v,int c)
{
	edge[Size].to=v;
	edge[Size].next=Adj[u];
	edge[Size].val=c;
	Adj[u]=Size++;
}


int num[maxn];
LL cnt[MX];

void DFS(int u,int fa,int val)
{
	for(int i=Adj[u];~i;i=edge[i].next)
	{
		int v=edge[i].to;
		int c=edge[i].val;
		if(v==fa) continue;
		num[v]=num[u]^c;
		DFS(v,u,num[v]);
	}
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d",&n);
		init();
		for(int i=0,a,b,c;i<n-1;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			Add_Edge(a,b,c); Add_Edge(b,a,c);
		}
		memset(cnt,0,sizeof(cnt));
		DFS(1,1,0);
		for(int i=1;i<=n;i++) cnt[num[i]]++;
		scanf("%d",&Q);
		while(Q--)
		{
			int x;
			scanf("%d",&x);
			LL ans=0;

			for(int i=1;i<=n;i++)
			{
				int u=num[i];
				int v=x^u;
				ans=ans+cnt[v];
			}
			if(x==0) ans+=n;
			printf("%lld\n",ans/2);
		}
	}
    
    return 0;
}