2015 多校联赛 ——HDU5416（异或）

 

CRB has a tree, whose vertices are labeled by 1, 2, …, N – 1 edges. Each edge has a weight.
For any two vertices v(possibly equal), ) is xor(exclusive-or) sum of weights of all edges on the path from v.
CRB’s task is for given s, to calculate the number of unordered pairs ) such that 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 c denoting an edge between 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 ≤ 5
1 ≤ Q ≤ 10
1 ≤ N
0 ≤ 5
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\n",ans);
        }
    }
    return 0;
}