AtCoder Grand Contest 012 D Colorful Balls

洛谷传送门

AtCoder 传送门

不错的题。bx Ender32k。

我们发现交换有传递性，那么我们能交换就连边，答案是 \(\prod \frac{(sz)!}{\prod c_i!}\)，其中 \(sz\) 为连通块大小，\(c_i\) 为这个连通块中第 \(i\) 种颜色出现次数。于是我们得到了一个 \(O(n^2)\) 的做法。

发现很多遍是无用的，考虑舍弃无用边。对于第一种连边，我们只与这种颜色的最小值的点连边；对于第二种连边，我们只与最小值最小和次小的点连边。然后就做完了。

code

// Problem: D - Colorful Balls
// Contest: AtCoder - AtCoder Grand Contest 012
// URL: https://atcoder.jp/contests/agc012/tasks/agc012_d
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<int, int> pii;

const int maxn = 200100;
const int N = 200000;
const ll mod = 1000000007;
const int inf = 0x3f3f3f3f;

inline ll qpow(ll b, ll p) {
	ll res = 1;
	while (p) {
		if (p & 1) {
			res = res * b % mod;
		}
		b = b * b % mod;
		p >>= 1;
	}
	return res;
}

int n, X, Y, a[maxn], b[maxn], p[maxn], stk[maxn], top, m, d[maxn];
pii c[maxn];
ll fac[maxn], ifac[maxn], ans = 1;
bool vis[maxn];
vector<int> G[maxn];

inline ll C(ll n, ll m) {
	if (n < m || n < 0 || m < 0) {
		return 0;
	} else {
		return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
	}
}

inline void init() {
	fac[0] = 1;
	for (int i = 1; i <= N; ++i) {
		fac[i] = fac[i - 1] * i % mod;
	}
	ifac[N] = qpow(fac[N], mod - 2);
	for (int i = N - 1; ~i; --i) {
		ifac[i] = ifac[i + 1] * (i + 1) % mod;
	}
}

void dfs(int u) {
	vis[u] = 1;
	ans = ans * ifac[m] % mod * fac[d[a[u]]] % mod;
	++d[a[u]];
	++m;
	ans = ans * fac[m] % mod * ifac[d[a[u]]] % mod;
	stk[++top] = a[u];
	for (int v : G[u]) {
		if (!vis[v]) {
			dfs(v);
		}
	}
}

void solve() {
	scanf("%d%d%d", &n, &X, &Y);
	for (int i = 1; i <= n; ++i) {
		c[i] = make_pair(inf, -1);
		p[i] = i;
	}
	for (int i = 1; i <= n; ++i) {
		scanf("%d%d", &a[i], &b[i]);
		c[a[i]] = min(c[a[i]], make_pair(b[i], i));
	}
	sort(p + 1, p + n + 1, [&](int x, int y) {
		return c[x] < c[y];
	});
	for (int i = 1; i <= n; ++i) {
		if (c[a[i]].fst + b[i] > X || c[a[i]].scd == i) {
			continue;
		}
		G[i].pb(c[a[i]].scd);
		G[c[a[i]].scd].pb(i);
	}
	if (n > 1 && c[p[2]].scd != -1) {
		for (int i = 1; i <= n; ++i) {
			for (int j = 1; j <= 2; ++j) {
				if (a[c[p[j]].scd] != a[i] && b[c[p[j]].scd] + b[i] <= Y) {
					G[i].pb(c[p[j]].scd);
					G[c[p[j]].scd].pb(i);
				}
			}
		}
	}
	for (int i = 1; i <= n; ++i) {
		if (vis[i]) {
			continue;
		}
		m = 0;
		dfs(i);
		while (top) {
			int x = stk[top--];
			d[x] = 0;
		}
	}
	printf("%lld\n", ans);
}

int main() {
	init();
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}