HDU 6280 From Tree to Graph（2018 湘潭邀请 E题，树的返祖边）

其实打返祖边就相当于$x$到祖先这一段点（不包括两端）答案都要减$1$.

然后每个点最多减$1$次$1$。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second

typedef long long LL;

const int N = 5e3 + 10;

bool c[N][N];
int  used[N];
int  father[N], d[N];
int  n, m, a, b, _x, _y;
int  ans;
vector <int> v[N];
vector <int> cnt;

void dfs(int x, int fa){
	father[x] = fa;
	for (auto u : v[x]){
		if (u == fa) continue;
		dfs(u, x);
	}
}

int main(){

	while (~scanf("%d%d%d%d%d%d", &n, &m, &a, &b, &_x, &_y)){
		rep(i, 0, n + 1) v[i].clear();

		rep(i, 1, n){
			rep(j, 1, n) c[i][j] = 0;
		}

		rep(i, 2, n){
			int x, y;
			scanf("%d%d", &x, &y);
			++x;
			++y;
			v[x].push_back(y);
			v[y].push_back(x);
		}

		dfs(1, 0);

		ans = 0;
		rep(i, 1, n){
			d[i] = (int)v[i].size();
			ans ^= d[i];
		}

		rep(i, 1, n) used[i] = 1;
		used[1] = 0;

		rep(i, 1, n){
			for (int j = i; j; j = father[j]){
				c[j][i] = 1;
			}
		}

		rep(i, 1, m){
			int x = _x, y = _y;
			_x = (a * x + b * y + ans) % n;
			_y = (b * x + a * y + ans) % n;

			x = _x + 1;
			y = _y + 1;

			printf("a = %d %d\n", x, y);

			cnt.clear();
			for (; x && !c[father[x]][y]; x = father[x]){
				cnt.push_back(x);
				if (father[x] == 1) continue;
				ans ^= d[father[x]];
				d[father[x]] -= used[x];
				used[x] = 0;
				ans ^= d[father[x]];
			}

			for (auto u : cnt) father[u] = x;
		}

		printf("%d %d\n", _x, _y);
	}

	return 0;
}