zoukankan      html  css  js  c++  java
  • HDU 4405 Aeroplane chess(概率DP)

    Aeroplane chess

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 4033 Accepted Submission(s): 2577


    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

    /*
    n个格子,m的特殊格子(可以在不掷骰子的情况下直接从xi飞到yi),求到终点的期望。
    求期望是从后往前推;
    */
    #include <algorithm>
    #include <iostream>
    #include <cmath>
    #include <cstdio>
    #include <cstring>
    #include <map>
    #include <vector>
    using namespace std;
    typedef long long LL;
    const LL N = 100000 + 10;
    const LL MOD = 1000;
    double dp[N];
    bool vis[N];
    vector<int>ve[N];
    int main()
    {
        int n, m;
        while(~scanf("%d%d", &n, &m)) {
            if(n + m == 0)
                break;
            int x, y, i, j, tmp;
            memset(dp, 0, sizeof(dp));
            memset(vis, false, sizeof(vis));
            for(i = 0; i <= n; i++)
                ve[i].clear();
            for(i = 0; i < m; i++) {
                scanf("%d%d", &x, &y);
                ve[y].push_back(x);
            }
            for(i = 0; i < ve[n].size(); i++) {
                tmp = ve[n][i];
                dp[tmp] = 0;
                vis[tmp] = true;
            }
            for(i = n - 1; i >= 0; i--) {
                if(!vis[i]) {
                    vis[i] = true;
                    for(j = i + 1; j <= i + 6; j++)
                        dp[i] += dp[j] / 6;
                    dp[i] += 1;
                }
                for(j = 0; j < ve[i].size(); j++) {
                    tmp = ve[i][j];
                    dp[tmp] = dp[i];
                    vis[tmp] = true;
                }
            }
            printf("%.4lf
    ", dp[0]);
        }
        return 0;
    }
  • 相关阅读:
    MiniGUI
    Android-在XML和Java代码中设置背景在不同状态的效果: <selector>/StateListDrawable
    URLEncoder.encode、URLDecoder.decode、escape、encodeURI、encodeURIComponent
    getDimension,getDimensionPixelOffset和getDimensionPixelSize的一点说明
    Android获取屏幕分辨率及DisplayMetrics简介
    细说Android事件传递机制(dispatchTouchEvent、onInterceptTouchEvent、onTouchEvent)
    Android
    Android坐标
    Android Sqlite IN, NOT IN syntax --- not int (?)
    TextView with SingleLine as true and Gravity as Center not passing the events to the ViewPager if it has a Click Event
  • 原文地址:https://www.cnblogs.com/yu0111/p/6776910.html
Copyright © 2011-2022 走看看