（HDU 4405） Aeroplane chess （概率DP）

最新推荐文章于 2021-03-30 14:32:48 发布

转载最新推荐文章于 2021-03-30 14:32:48 发布 · 92 阅读

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原文链接：http://www.cnblogs.com/brotherHaiNo1/p/8432550.html

本文介绍了一种计算飞机棋游戏中玩家达到终点所需的骰子投掷次数的期望值的方法。通过动态规划算法，考虑了普通移动和特殊飞行线路的影响，最终实现了对不同情况下的精确计算。

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

每次投一个色子，所以有6种不同的转移可能。

如果i位置不是一条航线的起点，那么：

dp[i]=(dp[i+1]+dp[i+2]+dp[i+3]+dp[i+4]+dp[i+5]+dp[i+6])/6+1;

否则(i到j有一条航线)：

dp[i]=dp[j](飞行可以不投色子)；

dp[0]就是答案；

#include<iostream>
#include<cstdio>
#include<vector>
#include<set>
#include<map>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#define LL long long
#define mod 1000000007
#define inf 0x3f3f3f3f

using namespace std;
int a[100010];
double dp[100010];

int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)
            break;
        memset(dp,0,sizeof(dp));
        memset(a,0,sizeof(a));
        for(int i=1;i<=m;i++)
        {
            int p,q;
            scanf("%d%d",&p,&q);
            a[p]=q;
        }
        for(int i=n-1;i>=0;i--)
        {
            if(a[i]!=0)
                dp[i]=dp[a[i]];
            else
            {
                double sum=1;
                for(int j=1;j<=6;j++)
                    sum+=(dp[i+j]/6);
                dp[i]=sum;
            }
        }
        printf("%.4lf\n",dp[0]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/brotherHaiNo1/p/8432550.html