HDU4405 Areoplane chess（期望dp）

Areoplane chess

传送门1
传送门2
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

---

题意

数轴上有$N+1$个点（编号$0-N$），一个人玩游戏，从$0$出发，当到达$N$或大于$N$的点则游戏结束。每次行动掷骰子一次，骰子编号$1-6$，掷到多少就向前走几步，这个数轴上还有些特殊点，可以从$Xi$到$Yi$。求总共投掷骰子次数的期望。

分析

期望dp.(期望dp一般逆推)
定义$dp[\,i\,]$表示在$i$时距离游戏结束还要投掷骰子次数的期望。
1. 对于可以直接飞的点，$dp[\,i\,]\!=\!dp[\,fly[\,i\,]].$
2. 不能飞的点，

$$dp[\,i\,]=\frac{\sum_{j=1}^6dp[\,i\!+\!j\,]}{6}+1$$
$$dp[\,i\,]\!=\!\frac{dp[\,i\!+\!1\,]\!+\!dp[\,i\!+\!2\,]\!+\!dp[\,i\!+\!3\,]+...\!+dp[\,i\!+\!6\,]}{6}+1$$

CODE

#include<cstdio>
#include<memory.h>
#define N 100005
#define FOR(i,a,b)  for(int i=(a),i##_END_=(b);i<=i##_END_;i++)
#define ROF(i,a,b)  for(int i=(a),i##_END_=(b);i>=i##_END_;i--)
double dp[N];
int fly[N];

int main() {
    int n,m;
    while(~scanf("%d%d",&n,&m)&&(n||m)) {
        memset(fly,-1,sizeof fly);
        FOR(i,1,m) {
            int a,b;
            scanf("%d%d",&a,&b);
            fly[a]=b;
        }
        memset(dp,0,sizeof dp);
        ROF(i,n-1,0) {
            if(fly[i]==-1) {
                FOR(j,i+1,i+6)dp[i]+=dp[j]/6.0;
                dp[i]++;
            } else dp[i]=dp[fly[i]];
        }
        printf("%.4lf\n",dp[0]);
    }
    return 0;
}