Josephina and RPG

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 929    Accepted Submission(s):
265
Special Judge

Problem Description
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
 
 /*                    dp[i+1][j]
dp[i][j]=
dp[i+1][num[i]]
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
const int ms=;
const int cms=;
const int MAXN=;
double dp[MAXN][cms];
double p[cms][cms];
int num[MAXN];
int main()
{
int i,j,k,t,n,m,cnt;
while(scanf("%d",&n)!=EOF)
{
for(cnt=,i=n;i>=(n-+);i--)
cnt*=i;
cnt/=;
for(i=;i<cnt;i++)
for(j=;j<cnt;j++)
scanf("%lf",&p[i][j]);
scanf("%d",&m);
for(i=;i<=m;i++)
scanf("%d",&num[i]);
for(i=;i<=cnt;i++)
dp[m+][i]=1.0;
for(i=m;i>;i--)
{
for(j=;j<cnt;j++)
{
dp[i][j]=p[j][num[i]]*max(dp[i+][j],dp[i+][num[i]]);
}
}
double ans=-1.0;
for(j=;j<cnt;j++)
if(ans<dp[][j])
ans=dp[][j];
printf("%.6lf\n",ans);
}
return ;
}

Josephina and RPG的更多相关文章

  1. 2013长沙赛区现场赛 J - Josephina and RPG

    J - Josephina and RPG Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I6 ...

  2. hdu4800 Josephina and RPG 解题报告

    Josephina and RPG Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others ...

  3. Josephina and RPG HDU - 4800

    A role-playing game (RPG and sometimes roleplaying game) is a game in which players assume the roles ...

  4. DP ZOJ 3735 Josephina and RPG

    题目传送门 题意:告诉你C(m,3)个队伍相互之间的胜率,然后要你依次对战n个AI队伍,首先任选一种队伍,然后战胜一个AI后可以选择替换成AI的队伍,也可以不换,问你最后最大的胜率是多少. 分析:dp ...

  5. hdu 4800 Josephina and RPG

    简单dp #include<cstdio> #define maxn 10005 #include<cstring> #include<algorithm> usi ...

  6. HDU 4800/zoj 3735 Josephina and RPG 2013 长沙现场赛J题

    第一年参加现场赛,比赛的时候就A了这一道,基本全场都A的签到题竟然A不出来,结果题目重现的时候1A,好受打击 ORZ..... 题目链接:http://acm.hdu.edu.cn/showprobl ...

  7. The 2013 ACM-ICPC Asia Changsha Regional Contest - J

    Josephina and RPG Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge A role-playin ...

  8. 2013 Asia Changsha Regional Contest---Josephina and RPG(DP)

    题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=4800 Problem Description A role-playing game (RPG and ...

  9. 【经典】C++&RPG对战游戏

    博文背景: 还记大二上学期的时候看的这个C++&RPG游戏(博主大一下学期自学的php,涵盖oop内容),一个外校的同学他们大一学的C++,大二初期C++实训要求做一个程序填空,就是这个 RP ...

随机推荐

  1. Core Java 学习笔记——2.基本数据类型&类型转换

    数据类型(8种基本类型:int/short/long/byte/float/double/char/boolean) 整型 int 4字节 -2 147 483 648~2 147 483 647 s ...

  2. CSS_网站配色参考方案

    http://www.cnblogs.com/QLeelulu/archive/2008/04/04/1136974.html   Shiny silver [#EEEEEE]       Reddi ...

  3. SRM 509 DIV1 500pt(DP)

    题目简述 给定一个字符串,可以对其进行修改,删除,增加操作,相应的操作有对应的花费,要求你用最小的花费把字符串变为回文串 题目做法 先搞一遍floyed把各种操作的最小花费求出来,然后就是类似编辑距离 ...

  4. http和数据库sql分析与窃听技术

    用tunnel,tunnel是一种技术称谓,将其放到真正的服务器和客户端之间.调试阶段可以使用webcream运行tomcat作为模拟的真正的服务器. 具体:用apache axis及其项目中的工具t ...

  5. google_protobuf数据类型

    要通信,必须有协议,否则双方无法理解对方的码流.在protobuf中,协议是由一系列的消息组成的.因此最重要的就是定义通信时使用到的消息格式. Protobuf消息定义 消息由至少一个字段组合而成,类 ...

  6. DelphiXE8新建AVD

    相关资料: 1.http://jingyan.baidu.com/article/ea24bc398576b3da62b33107.html

  7. spring注解使用

    一.各种注解方式 1.@Autowired注解(不推荐使用,建议使用@Resource) @Autowired可以对成员变量.方法和构造函数进行标注,来完成自动装配的工作.@Autowired的标注位 ...

  8. eclipse svn切换账号登陆问题

    1.当一个人有权限访问文件代码,而另一个账号无法访问该文件代码,要在eclipse上切换账号登陆有权限的账号时,eclipse会用缓存的账号,不会弹出从新输入新账号的窗口. 这样该怎么解决呢? 关闭e ...

  9. 斯特灵数 (Stirling数)

    @维基百科 在组合数学,Stirling数可指两类数,都是由18世纪数学家James Stirling提出的. 第一类 s(4,2)=11 第一类Stirling数是有正负的,其绝对值是个元素的项目分 ...

  10. [5] Zygote

    Android设备中的两大进程,如下图 1,由init进程创建的Daemon进程 2,由 Zygote进程创建的应用程序进程 什么是Zygote? zygote是“受精卵”的意思.在Android里, ...