标签：acm   hdu   

1
3
1 2 1
2 3 2
3
2
3
4

1
1
0
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.
 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <set>

using namespace std;
#define LL long long
#define N 100000 + 10

int 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;
}

/*

1
3
1 2 1
2 3 2
3
2
3
4

1
5
1 2 1
3 5 2
1 3 2
2 4 2
10
0
3
2

*/

hdu 5416

标签：acm   hdu   

原文地址：http://blog.csdn.net/dojintian/article/details/47834933