HDU 4405 Aeroplane chess（期望dp）

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5230    Accepted Submission(s): 3290

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

 

题意

一个1*n的网格，每次都可以以相同的概率从一个点往后跳1~6个点；另有m条路线，a->b表示从a只能直接跳到b。跳到大于等于n，后结束游戏，求结束游戏的步数期望。

分析

dp[i]表示从i点开始，到结束游戏的期望。

那么dp[n]=0，

dp[i]=(dp[j]+1)/6   j=i+1,i+2...i+6;

如果i处有飞行线，那么dp[i]=dp[f[i]]   f[i]是i点到达的点。

code

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

using namespace std;

int f[100100]; // 记录飞行路线
double dp[100100]; // dp[i]表示在i位置时，距离游戏结束还要投掷次数的期望

int main() {
    int n,m;
    while (~scanf("%d%d",&n,&m) && n+m) {
        for (int i=0; i<=n+5; ++i) f[i] = -1,dp[i] = 0.0;
        for (int a,b,i=1; i<=m; ++i) {
            scanf("%d%d",&a,&b);f[a] = b;
        }
        dp[n] = 0.0; // 在n点的期望步数是0
        for (int i=n-1; i>=0; --i) {
            if (f[i] == -1) {
                for (int j=1; j<=6; ++j) 
                    dp[i] += (dp[i+j]+1) / 6.0;
            }
            else dp[i] = dp[f[i]];
        }
        printf("%.4lf\n",dp[0]);
    }
    return 0;
}