（概率dp/期望）Aeroplane chess

最新推荐文章于 2021-03-30 14:32:48 发布

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最新推荐文章于 2021-03-30 14:32:48 发布

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本文探讨了一种飞行棋游戏的数学模型，棋盘由N+1格组成，玩家从0格开始，通过掷骰子移动。文章考虑了额外的M条航线，允许玩家在特定格子间瞬移，目标是计算到达N格所需的平均掷骰子次数。通过动态规划方法，文章详细解释了如何计算这一期望值。

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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.

题意即为飞行棋，0-n共n+1格，m条航线可以从xi-yi，没有航线的话可以扔骰子后移动到当前位置+骰子点数的位置，当位置>=n时结束，求扔骰子的次数期望
如果该地有航线dp[i]=dp[y[i]]直接继承期望，否则dp[i]=1/6*dp[i+1]+1/6*dp[i+2]+…+1/6*dp[i+6]+1;

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
typedef long long ll;
const int maxn = 100000 + 5;
double dp[maxn];
int air[maxn];
int main()
{
    int n, m, x, y;
    while(~scanf("%d%d", &n, &m)) {
        if(n == m && n == 0) break;
        memset(air, 0, sizeof(air));
        dp[n] = 0.0;
        for(int i = 1; i <= m; i++) {
            scanf("%d%d", &x, &y);
            air[x] = min(n, y);
        }
        for(int i = n - 1; i >= 0; i--) {
            if(air[i] > 0) dp[i] = dp[air[i]];
            else {
                dp[i] = 1.0;
                for(int j = 1; j <= 6; j++)
                    dp[i] += 1.0 / 6 * dp[(i + j > n) ? n : i + j];
            }
        }
        printf("%.4f\n", dp[0]);
    }
}