[CF895E] Eyes Closed(线段树,期望)

Desctiption

传送门:Portal

大致题意: 给你一个序列, 支持两种操作:

1. 1 l1 r1 l2 y2 在([l1, r1])随机选择一个数a, ([l2, r2]) 内随机选择一个数b, 交换a, b.
2. 2 l r 询问一个区间的期望.

[n leq 200000; a_i leq 1e9 ]

Solution

根据期望线性性,只需要维护出每个位置元素值的期望就可以.

区间操作期望问题的经典套路是维护出每个位置的期望.

考虑一个数字(a_i)

他有(frac{len - 1}{len}) 的概率保持原数, 否则,根据全期望公式, 它会变成(E(b_i) = sumfrac{b_i}{lenB})

那么有:

[A_i = frac{(lenA - 1)A_i + sum frac{b_i}{lenB}}{lenA} ]

(b_i)也同理.

所以我们只需要维护一个区间加/乘, 区间求和的线段树就可以了.

Code

#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for(int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
typedef long long LL;
typedef long double LD;
int read() {
    char ch = getchar();
    int x = 0, flag = 1;
    for (;!isdigit(ch); ch = getchar()) if (ch == '-') flag *= -1;
    for (;isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    return x * flag;
}
void write(int x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + 48);
}

const int Maxn = 200009;
int a[Maxn];

namespace SGMTtree {
	double tree[Maxn << 3], addTag[Maxn << 3], mulTag[Maxn << 3];
#define lc(x) ((x) << 1)
#define rc(x) (((x) << 1) | 1)
#define ls rt << 1, l, mid
#define rs (rt << 1) | 1, mid + 1, r
	void setAdd(int root, int l, int r, double v) {
		tree[root] += v * (r - l + 1.0); addTag[root] += v;
	}
	void setMul(int root, int l, int r, double v) {
		tree[root] *= v;
		mulTag[root] *= v; addTag[root] *= v;
	}
	void (*setTag)(int, int, int, double);
	void set(int p) {
		if(p == 1) setTag = setAdd;
		if(p == 2) setTag = setMul;
	}

	void pushdown(int rt, int l, int r) {
		int mid = (l + r) >> 1;
		if (mulTag[rt] != 1.0) {
			setMul(ls, mulTag[rt]), setMul(rs, mulTag[rt]);
			mulTag[rt] = 1;
		}
		if (addTag[rt] != 0.0) {
			setAdd(ls, addTag[rt]); setAdd(rs, addTag[rt]); 
			addTag[rt] = 0;
		}
	}
	void pushup(int rt) { tree[rt] = tree[lc(rt)] + tree[rc(rt)]; }

	void build(int rt, int l, int r) {
		addTag[rt] = 0, mulTag[rt] = 1;
		if (l == r) {
			tree[rt] = a[l];
			return ;
		}
		int mid = (l + r) >> 1;
		build(ls), build(rs);
		pushup(rt);
	}
	void modify(int rt, int l, int r, int p, int q, double v) {
		if (p > q) return ;
		if (p <= l && r <= q) {
			setTag(rt, l, r, v);
			return ;
		}
		int mid = (l + r) >> 1; pushdown(rt, l, r);
		if (q <= mid) modify(ls, p, q, v);
		else if (p > mid) modify(rs, p, q, v);
		else modify(ls, p, q, v), modify(rs, p, q, v);
		pushup(rt);
	}
	double query(int rt, int l, int r, int p, int q) {
		if (p <= l && r <= q) return tree[rt];
		int mid = (l + r) >> 1; pushdown(rt, l, r);
		if (q <= mid) return query(ls, p, q);
		else if (p > mid) return query(rs, p, q);
		else return query(ls, p, q) + query(rs, p, q);
	}
#undef lc
#undef rc
#undef ls
#undef rs
};

int n, q;

void init() {
	n = read(); q = read();
	rep (i, 1, n) a[i] = read();
	SGMTtree :: build(1, 1, n);
}

void solve() {
	rep (i, 1, q) {
		int opt = read();
		if (opt == 1) {
			int l1 = read(), r1 = read(), l2 = read(), r2 = read();
			double c = r1 - l1 + 1.0, d = r2 - l2 + 1.0, tmp1 = SGMTtree :: query(1, 1, n, l1, r1), tmp2 = SGMTtree :: query(1, 1, n, l2, r2);
			SGMTtree :: set(2), SGMTtree :: modify(1, 1, n, l1, r1, c - 1.0);
			SGMTtree :: set(2), SGMTtree :: modify(1, 1, n, l2, r2, d - 1.0);
			SGMTtree :: set(1), SGMTtree :: modify(1, 1, n, l1, r1, tmp2 / d);
			SGMTtree :: set(1), SGMTtree :: modify(1, 1, n, l2, r2, tmp1 / c);
			SGMTtree :: set(2), SGMTtree :: modify(1, 1, n, l1, r1, 1 / c);
			SGMTtree :: set(2), SGMTtree :: modify(1, 1, n, l2, r2, 1 / d);
		}
		if (opt == 2) {
			int l = read(), r = read();
			printf("%.7lf
", SGMTtree :: query(1, 1, n, l, r));
		}
	}
}

int main() {
//	freopen("CF895E.in", "r", stdin);
//	freopen("CF895E.out", "w", stdout);

	init();
	solve();

#ifdef Qrsikno
    debug("
Running time: %.3lf(s)
", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
    return 0;
}

调试:

线段树函数指针必须要用namespace.