HDU 4405 Aeroplane chess (概率DP & 期望）

最新推荐文章于 2021-03-30 14:32:48 发布

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本文介绍了一种计算飞机棋游戏中达到终点所需骰子投掷次数的期望值的方法。游戏在一个包含N+1个格子的地图上进行，玩家从第0个格子开始，通过掷骰子前进。地图上有M条飞行线路，允许玩家直接从一个格子跳到另一个格子，无需掷骰子。文章详细解释了如何通过动态规划算法计算从任意起始位置到达终点的期望投掷次数。

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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

思路：
当已走长度i>=n时，i再掷一次的期望为0，从n开始向下推，可知
n-1的期望=1/6*(n的期望)+1,
n-2的期望=1/6*(n的期望)+1/6*(n-1的期望)+1
同理，可推出i=0时的期望。

代码实现：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=100010;
const int Max=1010;
double dp[N];
int line[N];
int main()
{
    int m,n;
    while(scanf("%d%d",&n,&m),n!=0||m!=0)
    {
        memset(line,-1,sizeof(line));
        for(int i=1;i<=m;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            line[a]=b;
        }
        memset(dp,0,sizeof(dp));
        for(int i=n-1;i>=0;i--)
        {
            if(line[i]!=-1)
            {
                dp[i]=dp[line[i]];
            }
            else
            {
                for(int k=1;k<=6;k++)
                {
                    if(i+k<=n)
                        dp[i]+=(1.0/6*dp[i+k]);
                    else break;
                }
                dp[i]+=1;
            }
        }
        printf("%.4f\n",dp[0]);
    }
    return 0;
}