zoukankan      html  css  js  c++  java
  • Josephina and RPG HDU

    A role-playing game (RPG and sometimes roleplaying game) is a game in which players assume the roles of characters in a fictional setting. Players take responsibility for acting out these roles within a narrative, either through literal acting or through a process of structured decision-making or character development. 
    Recently, Josephina is busy playing a RPG named TX3. In this game, M characters are available to by selected by players. In the whole game, Josephina is most interested in the "Challenge Game" part. 
    The Challenge Game is a team play game. A challenger team is made up of three players, and the three characters used by players in the team are required to be different. At the beginning of the Challenge Game, the players can choose any characters combination as the start team. Then, they will fight with N AI teams one after another. There is a special rule in the Challenge Game: once the challenger team beat an AI team, they have a chance to change the current characters combination with the AI team. Anyway, the challenger team can insist on using the current team and ignore the exchange opportunity. Note that the players can only change the characters combination to the latest defeated AI team. The challenger team gets victory only if they beat all the AI teams. 
    Josephina is good at statistics, and she writes a table to record the winning rate between all different character combinations. She wants to know the maximum winning probability if she always chooses best strategy in the game. Can you help her? 

    Input

    There are multiple test cases. The first line of each test case is an integer M (3 ≤ M ≤ 10), which indicates the number of characters. The following is a matrix T whose size is R × R. R equals to C(M, 3). T(i, j) indicates the winning rate of team i when it is faced with team j. We guarantee that T(i, j) + T(j, i) = 1.0. All winning rates will retain two decimal places. An integer N (1 ≤ N ≤ 10000) is given next, which indicates the number of AI teams. The following line contains N integers which are the IDs (0-based) of the AI teams. The IDs can be duplicated.

    Output

    For each test case, please output the maximum winning probability if Josephina uses the best strategy in the game. For each answer, an absolute error not more than 1e-6 is acceptable.

     
    Sample Input

    4
    0.50 0.50 0.20 0.30
    0.50 0.50 0.90 0.40
    0.80 0.10 0.50 0.60
    0.70 0.60 0.40 0.50
    3
    0 1 2

    Sample Output

    0.378000

    题目大意就是: RPG游戏中有很多不同的角色,选择其中三个组成一支队伍,现已知每支队伍打败其他队伍的概率,然后给你一个需要打败的队伍的顺序,求打败所有队伍的最大的概率是多少。

    其实就是一道概率DP的题目,换或者不换。

    对需要打败的队伍进行枚举,考虑换了队伍和不换队伍两者概率大小,取最大的那种。

    // Asimple
    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <cstdlib>
    #include <string>
    #include <cstring>
    #include <stack>
    #include <set>
    #include <map>
    #include <cmath>
    #define INF 0x3f3f3f3f
    #define debug(a) cout<<#a<<" = "<<a<<endl
    #define test() cout<<"============"<<endl
    #define CLS(a,v) memset(a, v, sizeof(a))
    #define pas system("pause")
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    int dx[] = {-1,1,0,0,-1,-1,1,1}, dy[]={0,0,-1,1,-1,1,1,-1};
    const int maxn = 10000+5;
    const ll mod = 1000000007;
    ll n, m, T, len, cnt, num, ans, Max, k;
    //vector<int> a[maxn]; 
    double dp[maxn];
    double Map[400][400];
    int team[maxn];
    
    void input(){
        while( scanf("%lld", &n)!=EOF) {
            n = n*(n-1)*(n-2)/6;
            double res = 1.0;
            for(int i=0; i<n; i++) {
                for(int j=0; j<n; j++) {
                    scanf("%lf", &Map[i][j]);
                }
                dp[i] = 1.0;
            }
            scanf("%lld", &m);
            for(int i=0; i<m; i++) scanf("%d", &team[i]);
            for(int i=0; i<m; i++) {
                double te = 0.0;
                for(int j=0; j<n; j++) {
                    te = max(Map[j][team[i]]*dp[j], te);
                    dp[j] = dp[j] * Map[j][team[i]];
                }
                dp[team[i]] = te;
            }
            printf("%.6lf
    ", dp[team[m-1]]);
        }
        // pas;
    }
    
    int main() {
        input();
        return 0;
    } 
  • 相关阅读:
    20189221 2018-2019-2 《密码与安全新技术专题》第五周作业
    2018-2019-2 20189221 《网络攻防技术》第九周作业
    2018-2019-2 20189221 《网络攻防技术》第八周作业
    20189221 2018-2019-2 《密码与安全新技术专题》第四周作业
    2018-2019-2 20189221 《网络攻防技术》第七周作业
    2018-2019-2 20189221 《网络攻防技术》第六周作业
    20189221 2018-2019-2 《密码与安全新技术专题》第三周作业
    2018-2019-1 20189203《Linux内核原理与分析》第五周作业
    2018-2019-1 20189203《Linux内核原理与分析》第四周作业
    2018-2019-1 20189203《Linux内核原理与分析》第三周作业
  • 原文地址:https://www.cnblogs.com/Asimple/p/9025908.html
Copyright © 2011-2022 走看看