#期望dp#洛谷 6835 [Cnoi2020]线形生物

题目

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分析

设\(f[i]\)表示由点\(i\)走到点\(i+1\)的期望步数，

\(dp[i]\)表示由点1走到点\(i+1\)的期望步数，

那么\(dp\)为\(f\)的前缀和，最后答案为\(dp[n]\)，只用考虑怎么计算\(f\)

那么

\[f[i]=\frac{\sum_{y\in son}(dp[i-1]-dp[y-1])+f[i]}{deg[i]-1}+1
\]

化简得到

\[f[i]=\sum_{y\in son}(dp[i-1]-dp[y-1]+1)+1
\]

其中\(deg\)表示\(i\)的度数，包括\(i->i+1\)

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代码

#include <cstdio>
#include <cctype>
#include <algorithm>
#define rr register
using namespace std;
const int N=1000011,mod=998244353;
struct node{int y,next;}e[N];
int n,k,dp[N],f[N],as[N];
inline signed iut(){
	rr int ans=0; rr char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
	return ans;
}
inline signed mo(int x,int y){return x+y>=mod?x+y-mod:x+y;}
signed main(){
	iut(),n=iut(),k=iut();
	for (rr int i=1;i<=k;++i){
		rr int x=iut(),y=iut();
		e[i]=(node){y,as[x]},as[x]=i;
	}
	for (rr int i=1;i<=n;++i){
		for (rr int j=as[i];j;j=e[j].next)
		    f[i]=mo(mo(f[i],dp[i-1]),mod-dp[e[j].y-1]+1);
		dp[i]=mo(dp[i-1],f[i]=mo(f[i],1));
	}
	return !printf("%d",dp[n]);
}