HDU-4405 Aeroplane chess （概率DP）

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本文介绍了一道关于飞行棋的算法题，通过动态规划求解从起点到达终点所需的掷骰子次数的期望值。文章给出了详细的解题思路及完整代码实现。

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Aeroplane chess

http://acm.hdu.edu.cn/showproblem.php?pid=4405

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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

Sample Output

  
   1.1667
2.3441

题目大意：玩飞行棋，初始位置为0，每次可以掷骰子走与点数相同的步数，当走到的位置i>=n时，游戏结束，初次之外，还存在某些点可以不用掷骰子直接跳至后面的某一点出，求游戏结束时掷骰子次数的期望？

大致思路：设dp[i]表示从点i出走到结束游戏掷骰子次数的期望

①若点i出无法跳跃，则只能依靠掷骰子前进，则 dp[i]可以转化为六种状态：dp[i+1],dp[i+2],dp[i+3],dp[i+4],dp[i+5],dp[i+6]，概率均为：1/6

②若点i处可以跳跃，则可以转化为一种状态：dp[nxt[i]]，概率为：1

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int n,m,s,e;
int nxt[100009];//nxt[i]表示点i处可以直接跳至点nxt[i]
double dp[100009];//dp[i]表示从点i出走到结束游戏掷骰子次数的期望

int main() {
    while(scanf("%d%d",&n,&m),n!=0||m!=0) {
        dp[n]=dp[n+1]=dp[n+2]=dp[n+3]=dp[n+4]=dp[n+5]=0;//走到点n~n+5处还需掷骰子的次数的期望为0
        memset(nxt,-1,sizeof(nxt));
        while(m-->0) {
            scanf("%d%d",&s,&e);
            nxt[s]=e;
        }
        for(int i=n-1;i>=0;--i) {
            if(nxt[i]==-1) {//如果i点不能跳，则必定掷骰子
                dp[i]=0;
                for(int j=1;j<=6;++j) {
                    dp[i]+=dp[i+j]/6;
                }
                dp[i]+=1;
            }
            else {//否则i点直接跳
                dp[i]=dp[nxt[i]];
            }
        }
        printf("%.4lf\n",dp[0]);
    }
    return 0;
}