洛谷P3600 随机数生成器(期望dp 组合数)

题意

题目链接

Sol

一条重要的性质：如果某个区间覆盖了另一个区间，那么该区间是没有用的(不会对最大值做出贡献)

首先不难想到枚举最终的答案\(x\)。这时我们需要计算的是最大值恰好为\(x\)的概率。

发现不是很好搞，我们记\(P(x)\)表示最大值\(\leqslant x\)的概率，那么恰好为\(x\)的概率为\(P(x) - P(x - 1)\)

计算概率可以直接用定义：合法的方案/总方案(\(x^n\))

考虑如何计算合法方案：我们直接去枚举在询问区间中有多少个点\(\leqslant x\)，设\(g(j)\)表示选出\(j\)个\(\leqslant x\)的点且覆盖了所有询问区间的方案，显然这样可以做到不重不漏。

接下来直接dp计算\(g[j]\)，设\(f[i][j]\)表示覆盖了前\(i\)个位置，放了\(j\)个点的方案数，且\(i\)位置必须放的方案数。

\(f[i][j] = \sum_{fr[k] + 1 \leqslant fl[i]} f[k][j - 1]\)

\(fr[i]\)表示覆盖了\(i\)区间的最右区间的编号，\(fl[i]\)表示覆盖了\(i\)的最左区间的编号

转移的时候拿单调栈搞一下

复杂度\(O(n^2 logn)\)

#include<bits/stdc++.h>
#define Pair pair<LL, LL>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define LL long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 2001, mod =666623333, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '\n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, X, Q, flag[MAXN], fl[MAXN], fr[MAXN], g[MAXN], cnt, tmp[MAXN], f[MAXN][MAXN], que[MAXN], top, sum[MAXN];
int fp(int a, int p) {
	int base = 1;
	while(p) {
		if(p & 1) base = mul(base, a);
		a = mul(a, a); p >>= 1;
	}
	return base;
}
int inv(int x) {
	return fp(x, mod - 2);
}
int solve(int x) {//find the probability that max <= x
	int now = 0;
	for(int i = 1; i <= N; i++) add2(now, mul(g[i], mul(fp(x, i), fp(X - x, N - i))));//choose j point staticfiac
	return mul(now, inv(fp(X, N)));
}
Pair q[MAXN];
signed main() {
	Fin(a);
	N = read(); X = read(); Q = read();
	for(int i = 1; i <= Q; i++) q[i].fi = read(), q[i].se = read();
	for(int i = 1; i <= Q; i++)
		for(int j = 1; j <= Q; j++)
			if(!flag[j] && (i != j) && (q[i].fi <= q[j].fi && q[i].se >= q[j].se)) //不加!flag[j]会wa，因为可能左右端点都相同
				flag[i] = 1;
	for(int i = 1; i <= Q; i++) if(!flag[i]) q[++cnt] = q[i];
	Q = cnt;
	sort(q + 1, q + Q + 1);
	memset(fl, 0x3f, sizeof(fl));
	for(int i = 1; i <= Q; i++)
		for(int j = 1; j <= N; j++)
			if(q[i].fi <= j && q[i].se >= j) chmin(fl[j], i), chmax(fr[j], i);
			else if(q[i].se < j) chmax(fr[j], i);
	for(int i = 1; i <= N; i++) if(q[fr[i]].se < i) fl[i] = fr[i] + 1;

	/*
	for(int i = 1; i <= N; i++)
		for(int j = 1; j <= min(i, N); j++)
			for(int k = 0; k < i; k++)
				if(fr[k] + 1 >= fl[i]) add2(f[i][j], f[k][j - 1]);
	*/
	f[0][0] = 1;
	int l = 1, r = 1; que[1] = 0; sum[0] = 1;
	for(int i = 1; i <= N; i++) {
		while(l <= r && fr[que[l]] + 1 < fl[i]) {
			for(int j = 0; j <= N; j++)
				add2(sum[j], -f[que[l]][j]);
			l++;
		}
		for(int j = 1; j <= min(i, N); j++) f[i][j] = sum[j - 1];
		for(int j = 1; j <= N; j++) add2(sum[j], f[i][j]);
		que[++r] = i;
	}
	for(int i = 1; i <= N; i++)
		if(fr[i] == Q)
			for(int j = 1; j <= N; j++)
				add2(g[j], f[i][j]);
	LL ans = 0;
	for(int i = 1; i <= N; i++) tmp[i] = solve(i);
	for(int i = 1; i <= X; i++) add2(ans, mul(i, add(tmp[i], -tmp[i - 1])));
	cout << ans;
    return 0;
}
/*
32 4
*/