HDU 5606 tree

Problem Description

There is a tree(the tree is a connected graph which contains $n$ points and $n-1$ edges),the points are labeled from 1 to $n$,which edge has a weight from 0 to 1,for every point $i\in [1,n]$,you should find the number of the points which are closest to it,the clostest points can contain $i$ itself.

Input

the first line contains a number T,means T test cases.

for each test case,the first line is a nubmer $n$,means the number of the points,next n-1 lines,each line contains three numbers $u,v,w$,which shows an edge and its weight.

T\le 50,n\le 10^5,u,v\in[1,n],w\in[0,1]T≤50,n≤10​5​​,u,v∈[1,n],w∈[0,1]

Output

for each test case,you need to print the answer to each point.

in consideration of the large output,imagine ans_ians​i​​ is the answer to point $i$,you only need to output,ans_1~xor~ans_2~xor~ans_3..~ans_nans​1​​ xor ans​2​​ xor ans​3​​.. ans​n​​.

Sample Input

1
3
1 2 0
2 3 1

Sample Output

1 in the sample. ans_1=2ans​1​​=2 ans_2=2ans​2​​=2 ans_3=1ans​3​​=1 2~xor~2~xor~1=1
$2xor2xor1=1$,so you need to output 1

.
直接用并查集合并，最后直接求解即可。
#include<map>
#include<stack>
#include<queue>
#include<cmath>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;
int T, n, m, fa[maxn], x, y, fx, fy, z, ans[maxn];

int get(int x)
{
    if (fa[x] != x) fa[x] = get(fa[x]);
    return fa[x];
}

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) fa[i] = i, ans[i] = 0;
        for (int i = 1; i < n; i++)
        {
            scanf("%d%d%d", &x, &y, &z);
            if (z) continue;
            fx = get(x);    fy = get(y);
            if (fx == fy) continue;
            if (fx > fy) swap(fx, fy);
            fa[fy] = fx;
        }
        int t = 0;
        for (int i = 1; i <= n; i++) ans[get(i)]++;
        for (int i = 1; i <= n; i++) if (ans[i] & 1) t ^= ans[i];
        printf("%d\n", t);
    }
    return 0;
}