【概率dp】HDU - 4405 Aeroplane chess

最新推荐文章于 2021-01-19 20:52:36 发布

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本文介绍了一个飞机棋游戏的算法实现，通过动态规划计算玩家完成游戏所需的骰子投掷次数的期望值。游戏地图包含多个格子及飞行线路，玩家从起点开始，每轮投掷骰子前进相应的格数，直至到达终点。

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Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5705 Accepted Submission(s): 3562

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

#include<cstdio>
#include<cstring>

int main()
{
    int n,m;
    double dp[112345];
    int jumpto[112345];
    while(scanf("%d%d",&n,&m)!=EOF&&!(n==0&&m==0))
    {
        memset(dp,0,sizeof(dp));
        memset(jumpto,-1,sizeof(jumpto));
        for(int i = 0; i < m; i++)
        {
            int x, y;
            scanf("%d %d", &x, &y);
            jumpto[x] = y;
        }
        dp[n]=0;
        for(int i=n-1; i>=0; i--)
        {
            if(jumpto[i]!=-1)
                dp[i]=dp[jumpto[i]];
            else
            {
                for(int j=1; j<=6; j++)
                {
                    if(i+j>=n)
                        break;
                    dp[i]+=1.0/6.0*dp[i+j];
                }
                dp[i]++;
            }
        }
        printf("%.4f\n",dp[0]);
    }


}