zoukankan      html  css  js  c++  java
  • HDU

    Aeroplane chess

     
    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. 

    InputThere 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. 
    OutputFor 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




    飞行棋求步数期望。
    逆推求期望典型。dp[i]下标状态为所在位置,值为步数。
    n=3,m=0
     
    e[3]=0
    e[2]=1/6e[3]+1
    e[1]=1/6e[2]+1/6e[3]+1
    e[0]=1/6e[1]+1/6e[2]+1/6e[3]+1
     
    
    
    #include<bits/stdc++.h>
    #define MAX 100010
    using namespace std;
    typedef long long ll;
    
    int f[MAX];
    double dp[MAX];
    
    int main()
    {
        int t,n,m,i,j;
        int x,y;
        while(scanf("%d%d",&n,&m)&&n+m){
            memset(f,-1,sizeof(f));
            memset(dp,0,sizeof(dp));
            for(i=1;i<=m;i++){
                scanf("%d%d",&x,&y);
                f[x]=y;
            }
            dp[n]=0;
            for(i=n-1;i>=0;i--){
                if(f[i]>-1){
                    dp[i]+=dp[f[i]];
                }
                else{
                    for(j=1;j<=6;j++){
                        if(i+j>n) break;
                        dp[i]+=1.0/6*dp[i+j];
                    }
                    dp[i]++;
                }
            }
            printf("%.4f
    ",dp[0]);
        }
        return 0;
    }
    
    
    
    
    
  • 相关阅读:
    ArrayList与LinkedList区别
    ArrayList底层原理
    nginx启用https访问
    云服务器搭建 Nginx 静态网站
    在云服务器上(CentOS)上安装Node
    文本超出显示省略号CSS
    vue使用改变element-ui主题色
    vue中的select框的值动态绑定
    vue项目对axios的全局配置
    使用crypto-js对数据进行AES加密、解密
  • 原文地址:https://www.cnblogs.com/yzm10/p/9678088.html
Copyright © 2011-2022 走看看