HDOJ 4405 Aeroplane chess（期望DP）_hdoj 期望-优快云博客

本文介绍了一种计算飞行棋游戏中达到终点所需掷骰子次数期望值的方法。通过正向推概率、反向推期望的方式，利用动态规划定义dp数组来解决含有飞行通道的飞行棋游戏问题。

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Aeroplane chess

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

Sample Output

1.1667
2.3441

题目大意：有一个N + 1格子的飞行棋棋盘，现在飞机在0号格子，你可以掷骰子，骰子上的数字对应你的飞机移动的步数。棋盘上还有一些飞行通道，即可以从a点飞到b点而不需要掷骰子，如果b点是另一条飞行通道的起点，那么飞机可以继续飞行而不用掷骰子。求到达终点需要投掷骰子次数的期望。

解题思路：正向推概率，反向推期望。定义dp[i]为从i点到达n点所需要投掷骰子次数的期望。dp[i]可以由6个状态转移过来，用一个循环处理即可。另外需要一个数组记录飞行通道的终点。

代码如下：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;

const int maxn = 100005;
int n,m;
int path[maxn];
double dp[maxn];


int main()
{
    while(scanf("%d %d",&n,&m) != EOF && (n || m)){
        int x,y;
        memset(path,-1,sizeof(path));
        memset(dp,0,sizeof(dp));
        for(int i = 0;i < m;i++){
            scanf("%d %d",&x,&y);
            path[x] = y;
        }
        for(int i = n - 1;i >= 0;i--){

            if(path[i] != -1){
                dp[i] = dp[path[i]];
                continue;
            }
            for(int j = 1;j <= 6;j++){
                if(i + j <= n)
                    dp[i] += dp[i + j] / 6.0;
                else
                    break;
            }
            dp[i] += 1;
        }
        printf("%.4f\n",dp[0]);
    }
    return 0;
}