hdu 4405 飞行棋（dp）

最新推荐文章于 2021-01-19 20:52:36 发布

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最新推荐文章于 2021-01-19 20:52:36 发布

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本文介绍了一种解决飞行棋游戏期望掷骰子次数问题的方法，通过动态规划算法实现。详细阐述了算法步骤，包括初始化、状态转移方程及最终计算结果的过程。

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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

题目描述：
飞行棋，掷骰子有等概率得到1-6之间的数字n，然后可以走n步。在棋盘上有飞行点和降落点，走到飞行点时会自动跳到降落点，求从起点到终点的步数的期望。

思路：
dp，计算到终点的距离，则终点处的步数期望为0，一个点能到后面6个点的概率均为1/6，故能得出转移方程 E[i]=E[i+1]*1/6+...+E[i+6]*1/6+1；因为有飞行点的存在，从飞行点到降落点是不需要掷骰子的，所以此时飞行点的期望即降落点的期望。

代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAXN 100500
int path[MAXN];
int n,m;
double e[MAXN];
int main()
{
 int i,j,a,b;

 while(~scanf("%d%d",&n,&m))
 {
  if(!n&&!m)
   break;
  memset(path,0,sizeof(path));

  for(i=0;i<m;i++)
  {
   scanf("%d%d",&a,&b);
   path[a]=b;
  }

  e[n]=0;
  for(i=n-1;i>=0;i--)
  {
   if(path[i])
    e[i]=e[path[i]];
   else
   {
    e[i]=1;
    for(j=1;j<=6;j++)
    {
     if(i+j<=n)
     e[i]=e[i]+(1.0/6)*e[i+j];
     else
     break;
    }
   }

  }
  printf("%.4f\n",e[0]);
 }
return 0;
}