hdu 5416 CRB and Tree(2015 Multi-University Training Contest 10)

CRB and Tree

                                                            Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                           Total Submission(s): 79    Accepted Submission(s): 16

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

131 2 12 3 23234

 

Sample Output

110
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

 

   

   

  解题思路：

       首先对于从节点u到节点v的异或值等于u到根节点的异或值再异或v到根节点的异或值,这是因为a^b=a^c^c^b,

   

   于是可以dfs求出所有节点到根节点的异或值，接着就是求所有异或值为s的情况，我们枚举一个u到根节点的值x,

   

   则v到根节点值为s^x,根据dfs的结果可以直接找到，因为u,v是无序的，会出现x==s^x的情况，特殊考虑就可。

  代码:

#include <iostream>#include <cstring>#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;const int maxn=131072;struct EDGE{    int to,v,next;}edge[200010];int ne=0;int head[100010];int sum[200010];int n;void addedge(int s,int e,int v){    edge[ne].to=e;    edge[ne].next=head[s];    edge[ne].v=v;    head[s]=ne++;}void dfs(int now,int pre,int nows){    sum[nows]++;    for(int i=head[now];i!=-1;i=edge[i].next)    {        if(edge[i].to==pre) continue;        dfs(edge[i].to,now,nows^edge[i].v);    }}int main(){    int T,i;    cin>>T;    while(T--)    {        ne=0;        memset(head,-1,sizeof(head));        cin>>n;        for(i=0;i<n-1;i++)        {            int a,b,c;            scanf("%d %d %d",&a,&b,&c);            addedge(a,b,c);            addedge(b,a,c);        }        memset(sum,0,sizeof(sum));        dfs(1,0,0);        int q,s;        cin>>q;        while(q--)        {            long long ans1=0,ans2=0;            cin>>s;            for(i=0;i<131072;i++)            {                int x=i,y=s^i;                if(x!=y)                    ans1+=(1ll*sum[x]*sum[y]);                else                {                    ans1+=(1ll*sum[x]*(sum[x]-1));                    ans2+=1ll*sum[x];                }            }            cout<<ans1/2+ans2<<endl;        }    }}