Josephina and RPG

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 511    Accepted Submission(s): 139
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

坑点: 有可能超过c(m,3),忽略可A

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <iostream>
#define INF 10000005000
using namespace std; const int maxm=10500;
const int maxn=150; typedef long long LL; int c[20][20],a[maxm];
double dp[2][maxn];
double cc[maxn][maxn]; void init()
{
memset(c,0,sizeof(c));
c[0][0]=c[1][0]=c[1][1]=1;
for(int i=2;i<=15;i++){
c[i][0]=c[i][i]=1;
for(int j=1;j<i;j++) c[i][j]=(c[i-1][j]+c[i-1][j-1]);
}
} int main()
{
int N,n,M,cur;
double res;
init();
while(scanf("%d",&N)!=EOF){
n=c[N][3];
for(int i=0;i<n;i++) for(int j=0;j<n;j++) scanf("%lf",&cc[i][j]);
memset(dp,0,sizeof(dp));
scanf("%d",&M);
for(int i=0;i<M;i++) {scanf("%d",&a[i]);if(a[i]>n)a[i]=0;}//ATTENTION!
for(int i=0;i<n;i++) dp[0][i]=cc[i][a[0]];
cur=1;
for(int i=1;i<M;i++){
for(int j=0;j<n;j++) dp[cur][j]=0;
for(int j=0;j<n;j++){
dp[cur][j]=max(dp[cur][j],dp[cur^1][j]*cc[j][a[i]]);
dp[cur][a[i-1]]=max(dp[cur][a[i-1]],dp[cur^1][j]*cc[a[i-1]][a[i]]);
}
cur^=1;
}
res=0;
for(int i=0;i<n;i++) res=max(res,dp[cur^1][i]);
printf("%.6f\n",res);
}
}

hdu4800 Josephina and RPG 解题报告的更多相关文章

  1. CH Round #56 - 国庆节欢乐赛解题报告

    最近CH上的比赛很多,在此会全部写出解题报告,与大家交流一下解题方法与技巧. T1 魔幻森林 描述 Cortana来到了一片魔幻森林,这片森林可以被视作一个N*M的矩阵,矩阵中的每个位置上都长着一棵树 ...

  2. 二模13day1解题报告

    二模13day1解题报告 T1.发射站(station) N个发射站,每个发射站有高度hi,发射信号强度vi,每个发射站的信号只会被左和右第一个比他高的收到.现在求收到信号最强的发射站. 我用了时间复 ...

  3. BZOJ 1051 最受欢迎的牛 解题报告

    题目直接摆在这里! 1051: [HAOI2006]受欢迎的牛 Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 4438  Solved: 2353[S ...

  4. 习题:codevs 2822 爱在心中 解题报告

    这次的解题报告是有关tarjan算法的一道思维量比较大的题目(真的是原创文章,希望管理员不要再把文章移出首页). 这道题蒟蒻以前做过,但是今天由于要复习tarjan算法,于是就看到codevs分类强联 ...

  5. 习题:codevs 1035 火车停留解题报告

    本蒟蒻又来写解题报告了.这次的题目是codevs 1035 火车停留. 题目大意就是给m个火车的到达时间.停留时间和车载货物的价值,车站有n个车道,而火车停留一次车站就会从车载货物价值中获得1%的利润 ...

  6. 习题: codevs 2492 上帝造题的七分钟2 解题报告

    这道题是受到大犇MagHSK的启发我才得以想出来的,蒟蒻觉得自己的代码跟MagHSK大犇的代码完全比不上,所以这里蒟蒻就套用了MagHSK大犇的代码(大家可以关注下我的博客,友情链接就是大犇MagHS ...

  7. 习题:codevs 1519 过路费 解题报告

    今天拿了这道题目练练手,感觉自己代码能力又增强了不少: 我的思路跟别人可能不一样. 首先我们很容易就能看出,我们需要的边就是最小生成树算法kruskal算法求出来的边,其余的边都可以删掉,于是就有了这 ...

  8. NOIP2016提高组解题报告

    NOIP2016提高组解题报告 更正:NOIP day1 T2天天爱跑步 解题思路见代码. NOIP2016代码整合

  9. LeetCode 解题报告索引

    最近在准备找工作的算法题,刷刷LeetCode,以下是我的解题报告索引,每一题几乎都有详细的说明,供各位码农参考.根据我自己做的进度持续更新中......                        ...

随机推荐

  1. 认识epoll

    linux下的epoll(7)函数,其有着良好的就绪事件通知机制.Epoll 是被linux2.6开始引进的,但是不被其他的类UNIX系统支持,它提供了一种类似select或poll函数的机制:a. ...

  2. getContext,getApplicationContext和this有什么区别

    使用this, 说明当前类是context的子类,一般是activity application等使用getApplicationContext 取得的是当前app所使用的application,这在 ...

  3. sql逻辑查询语句的执行顺序

    SELECT语句关键字的定义顺序 SELECT DISTINCT <select_list> FROM <left_table> <join_type> JOIN ...

  4. cygwin下使用apt-cyg安装新软件

    1.获取  (记得先安装好git) git clone https://github.com/transcode-open/apt-cyg.git 2.安装apt-cyg cd apt-cyg chm ...

  5. Centos7.2 修改网卡名称

    查看ip [root@localhost network-scripts]# ip addr : lo: <LOOPBACK,UP,LOWER_UP> mtu qdisc noqueue ...

  6. POJ 2373 Dividing the Path (单调队列优化DP)题解

    思路: 设dp[i]为覆盖i所用的最小数量,那么dp[i] = min(dp[k] + 1),其中i - 2b <= k <= i -2a,所以可以手动开一个单调递增的队列,队首元素就是k ...

  7. 封装TeeChart控件

    public class MyChart { //字段 private TChart tChart; /// <summary> /// 构造函数,默认不是3D效果 /// </su ...

  8. OJ上 编译器 G++和C++的区别

    原文 :http://blog.csdn.net/febr2/article/details/52068357 编译时的差异: 编译器优化不同: 举个栗子: ①: a++ ②: ++a 从标准C的角度 ...

  9. c#进阶之Delegate

    委托是什么?答:委托是一种类型   等同与 一个class类,继承System.MulticastDelegate,但mult....gate是一个特殊类,不能够派生 委托的调用,如何去使用 1/委托 ...

  10. Ansible 操作windows

      1.主控端安装ansible         1) pip install ansible 2.主控端安装相关的包 pip install http://github.com/diyan/pywi ...