hdu-5416（多校2015）

CRB and Tree

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 587    Accepted Submission(s): 187

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.

题意：给一棵树，树上的边对应一个值，q个询问s，即f(u,v)，从u到v的路径上的边值异或操作，结果等于s的路径数。

思路：f(u,v)=f(1,u)^f(1,v);

           s^f(1,v)=f(1,u);

//#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<stdio.h>#include<math.h>#include <string>#include<string.h>#include<map>#include<queue>#include<set>#include<utility>#include<vector>#include<algorithm>#include<stdlib.h>using namespace std;#define eps 1e-8#define pii pair<int,int>#define inf 0x3f3f3f3f#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define ll long long int#define mod 1000000007#define maxn 100005#define maxm 1000005vector<pii> f[maxn];int ff[maxn];ll num[maxm];int t,n,q,x,y,z;void dfs(int x,int pre,int k){//求f(1,x)    ff[x]=k;    num[k]++;    for(int i=0;i<f[x].size();i++)    {        int v=f[x][i].first;        int vv=f[x][i].second;        if(v==pre) continue;        dfs(v,x,k^vv);    }}int main(){    rd(t);    while(t--){        rd(n);        for(int i=1;i<=n;i++) f[i].clear();        for(int i=1;i<n;i++)        {            scanf("%d%d%d",&x,&y,&z);            f[x].push_back(make_pair(y,z));            f[y].push_back(make_pair(x,z));        }        memset(num,0,sizeof(num));        dfs(1,-1,0);        rd(q);        for(int i=1;i<=q;i++){            rd(x);            ll res=0;            for(int j=1;j<=n;j++)            {                res+=num[x^ff[j]];                if((x^ff[j])==ff[j]) res++;            }            printf("%I64d\n",res/2);        }    }    return 0;}