HDU 4405 Aeroplane chess 概率dp入门

题目链接：HDU 4405 Aeroplane chess

题目

Problem Description

Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces are 1,2,3,4,5,6). When Hzz is at grid i and the dice number is x, he will moves to grid i+x. Hzz finishes the game when i+x is equal to or greater than N.

There are also M flight lines on the chess map. The i-th flight line can help Hzz fly from grid Xi to Yi (0<Xi<Yi<=N) without throwing the dice. If there is another flight line from Yi, Hzz can take the flight line continuously. It is granted that there is no two or more flight lines start from the same grid.

Please help Hzz calculate the expected dice throwing times to finish the game.

Input

There are multiple test cases.
Each test case contains several lines.
The first line contains two integers N(1≤N≤100000) and M(0≤M≤1000).
Then M lines follow, each line contains two integers Xi,Yi(1≤Xi<Yi≤N).
The input end with N=0, M=0.

Output

For each test case in the input, you should output a line indicating the expected dice throwing times. Output should be rounded to 4 digits after decimal point.

Sample Input

2 0
8 3
2 4
4 5
7 8
0 0

Sample Output

1.1667
2.3441

题目解析

算是期望dp里的基础题了 （其实就是用来水一篇博客的） 。
为了照顾一下新手 （本来这就是道给新手的题），这里说一下概率dp的一个基本公式$$E(x)=C_x+\sum E(x_i)$$其中$x_i$是所有$x$能到达的后续状态,$C_x$是x点自身的贡献。放在这题里，每个点的后续状态很明确，就是从$x+1$到$x+6$，如果有滑翔道的话，则直接转移到滑翔道所在地点。
上代码：

#include <cstdio>
#include <cstring>
#include <string>
using namespace std;

#define maxn 100000 + 10
double dp[maxn];
int to[maxn];
int m, n;
const double p = 1.0/6.0;

void solve()
{
	memset(to, 0, sizeof(to));
	memset(dp, 0, sizeof(dp));
	int a, b;
	for(int i = 1; i <= m; i++){
		scanf("%d%d", &a, &b);	
		if(a > b) swap(a, b);
		to[a] = b;	
	}
	for(int i = n - 1 ; i >= 0; i--){
		if(to[i] != 0) dp[i] = dp[to[i]];
		else{
			for(int j = 1; j <= 6; j++)
				dp[i] += dp[i+j]*p;
			dp[i] += 1;
		}
	}
	printf("%.4f\n", dp[0]);
}

int main()
{
	while(scanf("%d%d", &n, &m), n)
		solve();
}