Josephina and RPG
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
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?
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.
probability if Josephina uses the best strategy in the game. For each answer, an
absolute error not more than 1e-6 is acceptable.
/* 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
- 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 ...
- 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 ...
- Josephina and RPG HDU - 4800
A role-playing game (RPG and sometimes roleplaying game) is a game in which players assume the roles ...
- 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 ...
- hdu 4800 Josephina and RPG
简单dp #include<cstdio> #define maxn 10005 #include<cstring> #include<algorithm> usi ...
- 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 ...
- 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 ...
- 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 ...
- 【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
- WeChat Official Account Admin Platform Message API Guide
Keyword: WeChat Message API Text Image Location Link Event Music RichMedia Author: PondBay Studio[We ...
- VS20xx下项目开发目录管理方法
在VS20XX之后项目管理使用解决方案(solution)管理一个大的开发工程中多个项目(Project). 以下目录配置与工程名称无关,适用于一般的工程组织过程. 对于一般的生成可执行程序的工程,使 ...
- ACM1998
/* 魔方阵,古代又称“纵横图”,是指组成元素为自然数1.2…n的平方的n×n的方阵, 其中每个元素值都不相等,且每行.每列以及主.副对角线上各n个元素之和都相等. 输入一个奇数,实现奇数魔方阵. 附 ...
- Android Studio的安装使用记录[持续更新]
参考资料: Windows环境下Android Studio v1.0安装教程 http://ask.android-studio.org/?/article/9 1. 下载与安装 在http://w ...
- JAVA NIO 类库的异步通信框架netty和mina
Netty 和 Mina 我究竟该选择哪个? 根据我的经验,无论选择哪个,都是个正确的选择.两者各有千秋,Netty 在内存管理方面更胜一筹,综合性能也更优.但是,API 变更的管理和兼容性做的不是太 ...
- c函数习记
1,user groups 篇幅 the length of an article; fgetgrent(从指定的文件来读取组格式) 相关related functions;fgetpwent hea ...
- append some buttons to the standard datagrid pager bar
<script type="text/javascript"> $(function(){ var pager = $('#dg').datagrid('getP ...
- Codeforces Round #375 (Div. 2) ABCDE
A - The New Year: Meeting Friends 水 #include<iostream> #include<algorithm> using namespa ...
- ASP.NET MVC中Session以及处理方式
转载原地址 http://www.cnblogs.com/darrenji/p/3951065.html
- Lua学习笔记(一):搭建开发环境
Lua是一个小巧高效的解释型脚本语言,可以方便的嵌入到任意的语言中,很多应用程序.游戏使用LUA作为自己的嵌入式脚本语言,以此来实现可配置性.可扩展性.这其中包括魔兽世界.博德之门.愤怒的小鸟.VOC ...