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. 当rsync遇到非默认端口的ssh

    在使用rsync使用ssh协议,来同步远程文件的方法,rsync -zvrtopg -e ssh但是如果遇到ssh不是22端口的时候使用rsync -zvrtopg -e ‘ssh -p 端口’特别是 ...

  2. 初识JAVA(【面向对象】:pub/fri/pro/pri、封装/继承/多态、接口/抽象类、静态方法和抽象方法;泛型、垃圾回收机制、反射和RTTI)

    JAVA特点: 语法简单,学习容易 功能强大,适合各种应用开发:J2SE/J2ME/J2EE 面向对象,易扩展,易维护 容错机制好,在内存不够时仍能不崩溃.不死机 强大的网络应用功能 跨平台:JVM, ...

  3. Hadoop 删除节点步骤

    1.在hadoop1.1.1/conf 下新建文件 nn-excluded-list 并写入要删除的节点名称或者IP 一个节点 一行 如: mos5200app cmpaknwom rac7 2.分发 ...

  4. 记录一次Android交叉编译ffmpeg排查错误

    Android版本手机直播引擎中,引用了libvlc开源库.项目接过来,发现编译脚本中使用了很多用户名下的绝对路径.项目相关人离职,导致这个脚本实际上已经废掉.而且不知道相关路径下有没有其他文件和第三 ...

  5. Java缓存学习之一:缓存

    一.缓存 1.什么是缓存? 缓存是硬件,是CPU中的组件,CPU存取数据的速度非常的快,一秒钟能够存取.处理十亿条指令和数据(术语:CPU主频1G),而内存就慢很多,快的内存能够达到几十兆就不错了,可 ...

  6. CodeForces 689E Mike and Geometry Problem (离散化+组合数)

    Mike and Geometry Problem 题目链接: http://acm.hust.edu.cn/vjudge/contest/121333#problem/I Description M ...

  7. linux性能问题(CPU,内存,磁盘I/O,网络)

    一. CPU性能评估 1.vmstat [-V] [-n] [depay [count]] -V : 打印出版本信息,可选参数 -n : 在周期性循环输出时,头部信息仅显示一次 delay : 两次输 ...

  8. C++常用容器

    vector 顺序容器,和数组类似,可从尾部快速的插入和删除,可随机访问. vector的常用成员函数: #include<vector> std::vector<type> ...

  9. 开源的读取Excel文件组件-ExcelDataReader

    ExcelDataReader可以读取 Microsoft Excel 文件 ('97-2007),支持Windows  .Net Framework 2 +. Windows Mobile with ...

  10. wikioi 3116 高精度练习之加法

    题目描述 Description 给出两个正整数A和B,计算A+B的值.保证A和B的位数不超过500位. 输入描述 Input Description 读入两个用空格隔开的正整数 输出描述 Outpu ...