HDU 4405

简单的概率DP,求期望,逆推更好。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 100005
using namespace std;

double p[N];

struct e{
	int u,v;
	int next;
}edge[1010];
int head[N],tot;

void addedge(int u,int v){
	edge[tot].u=u;
	edge[tot].v=v;
	edge[tot].next=head[u];
	head[u]=tot++;
}

int findx(int p){
	int ret=p;
	for(int ei=head[p];ei!=-1;ei=edge[ei].next){
		ret=edge[ei].v;
	}
	return ret;
}

int main(){
	int n,m,u,v,pos;
	while(scanf("%d%d",&n,&m),n||m){
		tot=0;
		memset(head,-1,sizeof(head));
		for(int i=1;i<=m;i++){
			scanf("%d%d",&u,&v);
			addedge(u,v);
		}
		p[n]=0;
		for(int i=n-1;i>=0;i--){
			p[i]=0;
			if(head[i]!=-1){
				p[i]=p[findx(i)];
			}
			else{
				for(int k=1;k<=6;k++){
					pos=i+k>n?n:i+k;
					if(head[pos]!=-1){
						p[i]+=p[findx(pos)]/6.0;
					}
					else{
						p[i]+=p[pos]/6.0;
					}
				}
				p[i]+=1;
			}
		}
		printf("%.4lf
",p[0]);
	}
	return 0;
}

  

原文地址:https://www.cnblogs.com/jie-dcai/p/4101097.html