P6835 [Cnoi2020]线形生物

新年快乐！

by もや造

题意

原题面

有(n+1)个点,对于每个点(i(i≤n))都有一条连向(i+1)的边,形成一条链,并在其中加入(m)条返祖边

现在从1号节点出发，每次等概率的前往到一个相邻的节点，求走到第(n+1)个点的期望步数

(n,m≤10^6)

分析

设(E_{x→y})表示从(x)点走到(y)点的期望步数，(k_i)表示第(i)个点的返祖边条数

有转移：

$~~~~~E_{x→x+1} = dfrac{1}{k_x+1}×1+dfrac{1}{k_x+1}×sumlimits_{(x,y)∈E}E_{y→x+1}+1$ $= 1+dfrac{1}{k_x+1}×sumlimits_{(x,y)∈E}E_{y→x+1}$

根据期望的线性性质,有(E_{x→y} =sumlimits_{i=x}^{y-1}E_{i→i+1}),代回原式：

$E_{x→x+1} = 1+dfrac{1}{k_x+1}×sumlimits_{(x,y)∈E}sumlimits_{i=y}^{x-1}E_{i→i+1}+dfrac{k_x×E_{x→x+1}}{k_x+1}$ $E_{x→x+1} = 1+k_x+sumlimits_{(x,y)∈E}sumlimits_{i=y}^{x-1}E_{i→i+1}$

维护一下前缀和即可,(sumlimits_{i=x}^{y-1}E_{i→i+1})既为所求答案

时间复杂度(O(n+m))

(code)

//xcxc82 2021/2/9/20:50 
#include<bits/stdc++.h>
#define mo 998244353
using namespace std;
const int MAXN = 1000010;
inline int read(){
	int X=0; bool flag=1; char ch=getchar();
	while(ch<'0'||ch>'9') {if(ch=='-') flag=0; ch=getchar();}
	while(ch>='0'&&ch<='9') {X=(X<<1)+(X<<3)+ch-'0'; ch=getchar();}
	if(flag) return X;return ~(X-1);
}
struct edge{
	int to,from,next;
}e[MAXN<<2];
int head[MAXN<<2],cnt;
void add(int u,int v){
	e[++cnt].next = head[u];
	e[cnt].from = u;
	e[cnt].to = v;
	head[u] = cnt;
}
int id,n,m,out[MAXN],sum[MAXN],f[MAXN];
int main(){
    id = read(),n =read(),m = read();
	for(int i=1;i<=m;i++){
		int u = read(),v =read();
		add(u,v);
		out[u]++;
	}
	for(int u=1;u<=n;u++){
		f[u] = 1+out[u];
		for(int i=head[u];i;i=e[i].next){
			int v = e[i].to;
			f[u] = (f[u]+(sum[u-1]-sum[v-1]+mo)%mo)%mo;
		}
		sum[u] = (sum[u-1]+f[u])%mo;
	}
	printf("%d",sum[n]);
   return 0;
}