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CRB and Tree

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 607    Accepted Submission(s): 194

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

 

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#include <cstdio>#include <cstring>#include <iostream>#include <set>using namespace std;#define LL long long#define N 100000 + 10int n, q, s, T;int head[N], tot;int m[2 * N];int dist[N];struct edge{    int v, w, next;}e[2 * N];void init(){    tot = 0;    memset(head, -1, sizeof head);    memset(m, 0, sizeof m);}void adde(int u, int v, int w){    e[tot].v = v;    e[tot].w = w;    e[tot].next = head[u];    head[u] = tot++;}void dfs(int u, int fa, int val){    dist[u] = val;    m[val]++;    for(int i = head[u]; i != -1; i = e[i].next)    {        int v = e[i].v;        if(v == fa) continue;        dfs(v, u, val ^ e[i].w);    }}void solve(){    LL ans = 0;    int j ;    for(int i = 1; i <= n; i++)    {        j = dist[i] ^ s;        if(j == dist[i]) ans += m[j] - 1;        else ans += m[j];    }    ans /= 2;    if(s == 0) ans += n;    printf("%I64d\n", ans);}int main(){    scanf("%d", &T);    while(T--)    {        scanf("%d", &n);        init();        int u, v, w;        for(int i = 1; i < n; i++)        {            scanf("%d%d%d", &u, &v, &w);            adde(u, v, w);            adde(v, u, w);        }        dfs(1, -1, 0);        scanf("%d", &q);        for(int i = 0; i < q; i++)        {            scanf("%d", &s);            solve();        }    }    return 0;}/*131 2 12 3 23234151 2 13 5 21 3 22 4 210032*/