2015 多校联赛 ——HDU5416（异或）

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

题意：
f(u，v) =  s(从u->v的异或)中u，v可以有多少对

所以f(u，v)  = f(1 , u) ^ f(1 , v)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define N 100050
typedef long long ll;


int head[N], tot;
int num[2*N];
int dis[N];

struct edge
{
    int v, w, next;
} edg[2*N];

void add(int u, int v, int w)
{
    edg[tot].v = v;
    edg[tot].w = w;
    edg[tot].next = head[u];
    head[u] = tot++;
}

void dfs(int u, int fa, int val)
{
    dis[u] = val;
    num[val]++;         //记录val的个数
    for(int i = head[u]; ~i; i = edg[i].next)
    {
        int v = edg[i].v;
        if(v == fa)
            continue;
        dfs(v, u, val^edg[i].w);
    }
}
int main()
{
    int T,u,v,w,m,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        tot = 0;
        memset(head,-1,sizeof(head));
        memset(num,0,sizeof(num));
        for(int i = 1; i < n; i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }

        dfs(1,-1,0);
        scanf("%d", &m);
        int s;
        for(int i = 0; i < m; i++)
        {
            scanf("%d", &s);
            int temp;
            ll ans = 0;
            for(int i = 1;i <= n;i++)
            {
                temp = s^dis[i];
                if(temp == dis[i])      //除去本身的那个
                    ans += num[temp]- 1;
                else
                    ans += num[temp];
            }
            ans/=2;
            if(s == 0)   //加上自己与自己异或的情况
                ans += n;
            printf("%I64d
",ans);
        }
    }
    return 0;
}