2017 ICPC Asia Urumqi I. A Possible Tree 帶權並查集

題目鏈接：https://nanti.jisuanke.com/t/40520

題解：因爲他們都是聯通的且只有唯一路徑，所以不用管之前怎麼連的，直接按照他給的查詢，帶權並查集判斷即可

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int f[N], r[N];
int n, m;
int fath(int x) {
	if(x == f[x]) return f[x];
	else {
		int fx = f[x];
		int rt = fath(f[x]);
		f[x] = rt;
		r[x] = r[x] ^ r[fx];
		return f[x];
	}
}
int main() {
	int T;
	int flag;
	int x, y, z;
	int xx, yy;
	scanf("%d", &T);
	while(T--) {
		flag = 0;
		scanf("%d %d", &n, &m);
		for(int i = 1; i <= n; i++) f[i] = i, r[i] = 0;
		for(int i = 1; i < n; i++) scanf("%d %d", &x, &y);
		for(int i = 1; i <= m; i++) {
			scanf("%d %d %d", &x, &y, &z);
			xx = fath(x);
			yy = fath(y);
			if(xx == yy) {
				if((r[x] ^ r[y]) != z && !flag) flag = i - 1;
			} else {
				f[xx] = yy;
				r[xx] = (r[x] ^ z ^ r[y]);
			}
		}
		printf("%d\n", flag);
	}
	return 0;
}