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:
    C语言和指针-回顾02-const
    Linux内核学习-使用exec创建socket
    Archlinux安装和配置
    apt-get install failed
    Insmod module : operation not permitted
    5.2.5.用开发板来调试模块
    5.2.4.最简单的模块源码分析3
    5.2.3.最简单的模块源码分析2
    5.2.1.开启驱动开发之路
    总线,设备,驱动的关系
  • 原文地址:https://www.cnblogs.com/yzm10/p/9678088.html
Copyright © 2011-2022 走看看