zoukankan      html  css  js  c++  java
  • 概率dp HDU 4405

    Aeroplane chess
    Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
    Appoint description: 

    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:Hzz在玩一种游戏,在N+1个格的图上,初始在0处,每次掷一枚骰子,骰子有6个标有1,2,3,4,5,6的面。每次前进最上面的那个面上所标的数步。有些地方有道具,到达之后它能够使用道具直接到达yi处,而不用掷骰子。求到达终点掷骰子的期望。

    /*************************************************************************
        > File Name: t.cpp
        > Author: acvcla
        > Mail: acvcla@gmail.com 
        > Created Time: 2014年10月21日 星期二 21时33分55秒
     ************************************************************************/
    #include<iostream>
    #include<algorithm>
    #include<cstdio>
    #include<vector>
    #include<cstring>
    #include<map>
    #include<queue>
    #include<stack>
    #include<string>
    #include<cstdlib>
    #include<ctime>
    #include<set>
    #include<math.h>
    using namespace std;
    typedef long long LL;
    const int maxn = 1e5 + 10;
    #define rep(i,a,b) for(int i=(a);i<=(b);i++)
    #define pb push_back
    double dp[maxn];
    int y[maxn],n,m;
    int main(int argc, char const *argv[])
    {
    	while(~scanf("%d%d",&n,&m)){
    		if(!n&&!m){
    			return 0;
    		}
    		n++;
    		memset(dp,0,sizeof dp);
    		memset(y,0,sizeof(y));
    		int x,to;
    		for(int i=1;i<=m;i++){
    			scanf("%d%d",&x,&to);
    			y[x+1]=to+1;
    		}
    		for(int i=n-1;i>=1;i--){
    			double t=0;
    			for(int j=1;j<=6;j++)t+=dp[i+j]/6;
    			if(y[i]){
    				t=dp[y[i]]-1;
    			}
    			dp[i]=t+1;
    		}
    		printf("%.4f
    ",dp[1]);
    	}
    	return 0;
    }




  • 相关阅读:
    TSINGSEE青犀视频AI智能识别功能开发如何通过GPU实现加速识别?
    沉浸式音视频互动要通过什么技术来实现?
    VR和AI进一步发展融合,智能视频的未来将会如何变革?
    什么样的安防摄像机才算是智能安防摄像机?
    编译EasyRTC新版本采用ProtocolBuffer(pb)接收不同类型数据如何判断?
    浅谈国内安防监控视频平台的未来发展和机遇
    摄像头接入EasyNVR和EasyCVR后视频流交互的区别在哪?
    新冠变异毒株来势汹汹,企业如何做好线上转型?
    推荐领域经典算法原理回顾 (LR / FMs / DT / GBDT / XGBoost)
    String字符串常量池
  • 原文地址:https://www.cnblogs.com/wzzkaifa/p/6844010.html
Copyright © 2011-2022 走看看