BestCoder Round #68 (div.2)

并查集 1002 tree

题意:中文题面

分析:(官方题解)把每条边权是1的边断开,发现每个点离他最近的点个数就是他所在的连通块大小.

开一个并查集,每次读到边权是0的边就合并.最后Ansi=size[findset(i)],sizeAns_i=size[findset(i)],sizeAns​i​​=size[findset(i)],size表示每个并查集根的size.

其实DFS也能过.

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

typedef long long ll;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;
struct DSU	{
	int rt[N], sz[N];
	void clear(void)	{
		memset (rt, -1, sizeof (rt));
		memset (sz, 0, sizeof (sz));
	}
	int Find(int x)	{
		return rt[x] == -1 ? x : rt[x] = Find (rt[x]);
	}
	void Union(int x, int y)	{
		x = Find (x);	y = Find (y);
		if (x == y)	return ;
		rt[y] = x;
		sz[x] += sz[y] + 1;
	}
}dsu;
int n;

int main(void)  {
    int T;  scanf ("%d", &T);
    while (T--) {
		dsu.clear ();
        scanf ("%d", &n);
        for (int u, v, w, i=1; i<n; ++i)   {
            scanf ("%d%d%d", &u, &v, &w);
            if (!w)	{
				dsu.Union (u, v);
			}
        }
		int ans = 0;
		for (int i=1; i<=n; ++i)	{
			ans ^= (dsu.sz[dsu.Find (i)] + 1);
		}
        printf ("%d
", ans);
    }

    return 0;
}