HDU 4405 Aeroplane chess （概率DP）——2012 ACM/ICPC Asia Regional Jinhua Online

传送门

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3520    Accepted Submission(s): 2232

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

题目大意：

有一个人在玩飞行棋，现在你开始的时候站在 0 点，终点是 n 点，给你一个色子（6个面），你掷色子得到几点就往前移动几步，但是这个飞行棋还有传送功能，可以从 x 点传送到 y 点，现在让你求从 0 点到 n 点置的色子的次数的期望（注意是次数的期望）。

解题思路：

其实这个题目很容易得到转移方程，现在设 DP[i] 表示的是 i 点到 n 点的次数的期望，那么可以得到方程就是 DP[i]=1+16∑6j=1DP[i+j]，那么我们现在在解决掉传送的问题，这个题目就解决了，从 x 传送到 y ，那么 DP[x]==DP[y]，所以我们可以用一个数组记录一下 vis[x]=y,所以 dp[x]==dp[vis[x]]，然后就可以从后往前循环，最终输出 DP[0] 就ok了。。

My Code：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <cmath>using namespace std;const int MAXN = 1e5+5;int vis[MAXN];double dp[MAXN];int main(){    ///freopen("in.txt","r",stdin);    int n, m, x, y;    while(~scanf("%d%d",&n,&m)){        if(n==0 && m==0)            break;        memset(vis, 0, sizeof(vis));        for(int i=1; i<=m; i++){            scanf("%d%d",&x,&y);            vis[x] = y;        }        dp[n] = 0;        for(int i=n-1; i>=0; i--){            if(!vis[i]){                double sum = 0;                for(int j=1; j<=6; j++)                    if(i+j < n)                        sum += dp[i+j];                dp[i] = sum/6+1;            }            else                dp[i] = dp[vis[i]];        }        printf("%.4f\n",dp[0]);    }    return 0;}