有nn个粒子位于一个数轴上，第ii个粒子都有pip_i的概率往右移动，否则往左移动，并且初始位于坐标xix_i处，其移动速度为viv_i。问粒子第一次发生碰撞的期望时间，保证不存在坐标相同的两个粒子

n≤105,−109≤x≤109,1≤v≤106n \leq 10^5, -10^9 \leq x \leq 10^9, 1 \leq v \leq 10^6

CF1287F

Solution

能够发生第一次碰撞的粒子对，一开始肯定是相邻的，所以我们可以处理出所有可能会发生的碰撞的时间。

考虑按时间排序后的第ii种碰撞的情况，其发生的概率为P(P(前i−1i-1种碰撞均未发生且第ii种碰撞发生)=P() = P(前i−1i-1种碰撞均未发生)−P() - P(前ii种碰撞均未发生))

考虑对所有的时刻进行 dpdp，设dpi,0/1dp_{i,0/1}表示考虑了前ii个点，且第ii个点往左/右时，在当前时刻之前没有发生碰撞的概率

相当于是在对于时间维扫描线，随着时间的推移，需要动态禁止一些转移途径，直接动态DP即可

具体实现可以看代码

时间复杂度：O(n×logn×23)O(n \times \log n \times 2^3)

Code

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#include <iostream>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <climits>
#include <queue>
#include <set>
#include <cmath>
#include <map>
#include <bitset>
#include <fstream>
#include <tr1/unordered_map>
#include <assert.h>

#define x first
#define y second
#define y0 Y0
#define y1 Y1
#define mp make_pair
#define pb push_back
#define DEBUG(x) cout << #x << " = " << x << endl;

using namespace std;

typedef long long LL;
typedef long double LD;
typedef pair <int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
	T sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

inline void proc_status ()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int MAXN = 1e5;
const int MOD = 998244353;

namespace MATH
{
	inline void ADD (int &a, int b) { if ((a += b) >= MOD) a -= MOD; }

	inline int Pow (int a, int b)
	{
		int ans = 1;
		for (int i = b; i; i >>= 1, a = (LL) a * a % MOD) if (i & 1) ans = (LL) ans * a % MOD;
		return ans;
	}
}

using namespace MATH;

int N;
struct info
{
	int x, v, p;
	
	inline void read () { x = ::read<int>(), v = ::read<int>(), p = (LL) ::read<int>() * Pow (100, MOD - 2) % MOD; }

} A[MAXN + 5];

struct matrix // 右1左0
{
	int A[2][2];

	inline matrix () { memset (A, 0, sizeof A); }

	inline int* operator [] (const int &x) { return A[x]; }

	inline matrix operator * (const matrix &rhs) const
	{
		matrix now;
		for (int i = 0; i < 2; ++i)
			for (int k = 0; k < 2; ++k) if (A[i][k])
				for (int j = 0; j < 2; ++j)
					ADD (now.A[i][j], (LL) A[i][k] * rhs.A[k][j] % MOD);
		return now;
	}
};

namespace SEG
{
#define mid ((l + r) >> 1)
#define ls o << 1
#define rs o << 1 | 1
#define lson ls, l, mid
#define rson rs, mid + 1, r
	const int MAX_NODE = MAXN * 4;

	matrix node[MAX_NODE + 5];

	inline void push_up (int o) { node[o] = node[ls] * node[rs]; }

	inline void build (int o, int l, int r)
	{
		if (l == r)
		{
			int p = A[l].p;
			node[o][0][0] = node[o][1][0] = (1 - p + MOD) % MOD;
			node[o][0][1] = node[o][1][1] = p;
			return ;
		}
		build (lson), build (rson);
		push_up (o);
	}

	inline void modify (int o, int l, int r, int x, int s0, int s1)
	{
		if (l == r) { node[o][s0][s1] = 0; return ; }
		if (x <= mid) modify (lson, x, s0, s1);
		else modify (rson, x, s0, s1);
		push_up (o);
	}

	inline int query () { return (node[1][0][0] + node[1][0][1]) % MOD; }

#undef mid
}

struct point
{
	int x, s0, s1, t;
	LD tt;

	inline point (int _x = 0, int _s0 = 0, int _s1 = 0, int _t = 0, LD _tt = 0) { x = _x, s0 = _s0, s1 = _s1, t = _t, tt = _tt; }
};

inline int cmp_t (point a, point b) { return a.tt < b.tt; }

vector <point> vec;

inline void Init ()
{
	for (int i = 1; i < N; ++i)
	{
		info l = A[i], r = A[i + 1];
		if (l.p != 0 && r.p != 1) 
			vec.pb (point (i + 1, 1, 0, (LL) (r.x - l.x) * Pow (l.v + r.v, MOD - 2) % MOD, (LD) (r.x - l.x) / (l.v + r.v)));
		if (l.p != 0 && r.p != 0 && l.v > r.v) 
			vec.pb (point (i + 1, 1, 1, (LL) (r.x - l.x) * Pow (l.v - r.v, MOD - 2) % MOD, (LD) (r.x - l.x) / (l.v - r.v)));
		if (l.p != 1 && r.p != 1 && l.v < r.v) 
			vec.pb (point (i + 1, 0, 0, (LL) (r.x - l.x) * Pow (r.v - l.v, MOD - 2) % MOD, (LD) (r.x - l.x) / (r.v - l.v)));
	}

	sort (vec.begin(), vec.end(), cmp_t);

	SEG :: build (1, 1, N);
}

inline void Solve ()
{
	if (N == 1) return void (puts("0"));
	Init ();

	int ans = 0, lst = 1;
	for (int i = 0; i < vec.size(); ++i)
	{
		int x = vec[i].x, s0 = vec[i].s0, s1 = vec[i].s1, t = vec[i].t;
		SEG :: modify (1, 1, N, x, s0, s1);
		int p = (lst - SEG :: query () + MOD) % MOD;
		ADD (ans, (LL) p * t % MOD);
		lst = SEG :: query ();
	}

	cout << ans << endl;
}

inline void Input ()
{
	N = read<int>();
	for (int i = 1; i <= N; ++i) A[i].read();
}

int main ()
{

#ifdef hk_cnyali
	freopen("F.in", "r", stdin);
	freopen("F.out", "w", stdout);
#endif

	Input ();
	Solve ();

	return 0;
}