WC模拟（1.8） T3 送你一朵圣诞树

送你一朵圣诞树

题目背景：

1.8 WC模拟T3

分析：并查集 + 堆 + 贪心

 

对于全局最小的那个点，选择了它的父亲之后，就一定会选择它，所以我们直接合并它和它的父亲，那么对于一个点集i，记录ti为点集中的点的个数，Si为点集中的点的权值和，那么，对于两个点集，如果点集i优于点集j，那么t_i * s_j > t_j * s_i，那么s_i /t_i < s_j / t_j，那么直接并查集 + 堆，每次合并当前块和父亲的块就可以了。

 

Source:

/*
	created by scarlyw
*/
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cctype>
#include <vector>
#include <set>
#include <queue>
#include <ctime>
#include <bitset>

inline char read() {
	static const int IN_LEN = 1024 * 1024;
	static char buf[IN_LEN], *s, *t;
	if (s == t) {
		t = (s = buf) + fread(buf, 1, IN_LEN, stdin);
		if (s == t) return -1;
	}
	return *s++;
}

///*
template<class T>
inline void R(T &x) {
	static char c;
	static bool iosig;
	for (c = read(), iosig = false; !isdigit(c); c = read()) {
		if (c == -1) return ;
		if (c == '-') iosig = true;	
	}
	for (x = 0; isdigit(c); c = read()) 
		x = ((x << 2) + x << 1) + (c ^ '0');
	if (iosig) x = -x;
}
//*/

const int OUT_LEN = 1024 * 1024;
char obuf[OUT_LEN], *oh = obuf;
inline void write_char(char c) {
	if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;
	*oh++ = c;
}

template<class T>
inline void W(T x) {
	static int buf[30], cnt;
	if (x == 0) write_char('0');
	else {
		if (x < 0) write_char('-'), x = -x;
		for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
		while (cnt) write_char(buf[cnt--]);
	}
}

inline void flush() {
	fwrite(obuf, 1, oh - obuf, stdout);
}

/*
template<class T>
inline void R(T &x) {
	static char c;
	static bool iosig;
	for (c = getchar(), iosig = false; !isdigit(c); c = getchar())
		if (c == '-') iosig = true;	
	for (x = 0; isdigit(c); c = getchar()) 
		x = ((x << 2) + x << 1) + (c ^ '0');
	if (iosig) x = -x;
}
//*/

const int MAXN = 30000 + 10;

struct node {
	int to, s, t;
	node(int to = 0, int s = 0, int t = 0) : to(to), s(s), t(t) {}
	inline bool operator < (const node &a) const {
		return (double)s / t > (double)a.s / a.t;
	}
} ;

std::priority_queue<node> q;
std::vector<int> edge[MAXN];

int n, x, y, val;
int w[MAXN], deg[MAXN], father[MAXN], sum[MAXN], size[MAXN], fa[MAXN];
bool able[MAXN];

inline void add_edge(int x, int y) {
	edge[x].push_back(y), edge[y].push_back(x);
}

inline void read_in() {
	R(n);
	for (int i = 1; i < n; ++i) R(x), R(y), add_edge(x, y);
	for (int i = 1; i <= n; ++i) R(w[i]), R(able[i]);
}

inline void dfs(int cur, int fa) {
	deg[cur] = 0;
	for (int p = 0; p < edge[cur].size(); ++p) {
		int v = edge[cur][p];
		if (v != fa) father[v] = cur, dfs(v, cur), deg[cur]++;
	}
}

inline int get_father(int x) {
	return (fa[x] == x) ? (x) : (fa[x] = get_father(fa[x]));
}

inline void solve() {
	long long ans = 0;
	for (int i = 1; i <= n; ++i) {
		if (able[i]) {
			dfs(i, 0);
			long long ret = 0;
			while (!q.empty()) q.pop();
			for (int j = 1; j <= n; ++j) {
				if (j != i) q.push(node(j, w[j], 1));
				sum[j] = w[j], size[j] = 1, ret += w[j], fa[j] = j;
			}
			while (!q.empty()) {
				node temp = q.top();
				int cur = q.top().to;
				q.pop();
				if (get_father(cur) != cur || size[cur] != temp.t) continue ;
				int fa = get_father(father[cur]);
//				std::cout << "cur == " << cur << " " << fa << '\n';
//				std::cout << size[cur] << " " << sum[cur] << " " << size[fa] << " " << sum[fa] << '\n';
				ret += (long long)size[fa] * sum[cur], ::fa[cur] = fa;
				size[fa] += size[cur], sum[fa] += sum[cur];
				if (fa != i) q.push(node(fa, sum[fa], size[fa]));
			}
			ans = std::max(ans, ret);
		}
	}
	std::cout << ans;
}

int main() {
	freopen("xmastree2.in", "r", stdin);
	freopen("xmastree2.out", "w", stdout);
	read_in();
	solve();
	return 0;
}