hdu 4405 Aeroplane chess 概率dp入门题

本文探讨了一个掷骰子游戏的优化问题,玩家通过掷骰子前进,利用飞行线可以跳过某些格子。文章详细介绍了如何通过动态规划计算最优路径,以及如何实现飞行线的合理运用,以最小化掷骰子次数的期望值。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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 08 32 44 57 80 0

Sample Output

1.16672.3441

 

题目大意:有n+1个格子排成一排,编号从0到n,并且在某些格子能直接跳到某个格子。

现在掷骰子,掷到多少就走几步,求掷骰子次数的期望。

 

期望:$E(x) = \sum  {p_i*x_i}$

 

定义dp[i]是从i到达终点的掷骰子期望次数,那么$$dp[i] = \sum_{j=1}^6 {(dp[i+j]+1)/6}$$。

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
using namespace std;
const int maxn = 100010;
int jump[maxn];
double dp[maxn];
int main()
{
    int n,m,a,b;
    while(scanf("%d%d",&n,&m)!=EOF&&n) {
        memset(jump,0,sizeof(jump));
        memset(dp,0,sizeof(dp));
        while(m--) {
            scanf("%d%d",&a,&b);
            jump[a] = b;
        }
        dp[n] = 0;
        for(int i=n-1;i>=0;i--) {
            if(jump[i]) dp[i] = dp[jump[i]];
            else
            for(int j=1;j<=6;j++) {
                dp[i] += (dp[i+j]+1)/6;
            }
        }
        printf("%.4f\n",dp[0]);
    }
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值