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

131 2 02 3 1

Sample Output

1in the sample.ans_1=2ans​1​​=2ans_2=2ans​2​​=2ans_3=1ans​3​​=12~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;}