题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4800

题面：

Josephina and RPG

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

Source

2013 Asia Changsha Regional Contest

解题：

每三个人组成一只队伍，所以R的大小为C（3,n）。给出的数组是某种组合对应某种组合的获胜几率。开始可以任选一种组合。dp[i][j]代表的是，i这个位置当前选用了j这个组合的成功率。由常规的i-1推到i似乎很难推，因为并不知道之前使用的是什么或者说之前是否进行了替换。此时就要进行转换，由i推到i+1。

dp[i+1][j]=max(dp[i+1][j],dp[i][j]*rate[j][rival[i+1]]);
dp[i+1][rival[i]]=max(dp[i+1][rival[i]],dp[i][j]*rate[rival[i]][rival[i+1]]);

下一位置的dp[i+1][j]的值为其原先值，和在i+1位置保持用j的成功率的最大值。

下一位置的dp[i+1][rival[i]]的值为其原先值,和在i+1位置用上一轮打败的对手的成功率的最大值。

答案即为遍历最后一个位置取所有组合中最大的一个。

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;
double rate[260][260],dp[10010][260],ans;
int rival[10010];
double max(double a,double b)
{
	return a>b?a:b;
}
int C(int a,int b)
{
	int res=1;
	for(int i=1;i<=a;i++)
		res=res*(b-i+1)/i;
	return res;
}
int main()
{
    int n,t;
	while(scanf("%d",&n)!=EOF)
	{
	  n=C(3,n);
	  ans=0.0;
      for(int i=0;i<n;i++)
		for(int j=0;j<n;j++)
			scanf("%lf",&rate[i][j]);
      scanf("%d",&t);
	  for(int i=0;i<t;i++)
		  scanf("%d",&rival[i]);
	  for(int i=0;i<t;i++)
		  for(int j=0;j<n;j++)
			  dp[i][j]=0.0;
      for(int i=0;i<n;i++)
		  dp[0][i]=rate[i][rival[0]];
	  for(int i=0;i<t-1;i++)
	  {
		  for(int j=0;j<n;j++)
		  {
			  dp[i+1][j]=max(dp[i+1][j],dp[i][j]*rate[j][rival[i+1]]);
			  dp[i+1][rival[i]]=max(dp[i+1][rival[i]],dp[i][j]*rate[rival[i]][rival[i+1]]);
		  }
	  }
	  for(int i=0;i<n;i++)
		  ans=max(ans,dp[t-1][i]);
	  printf("%.6lf\n",ans);
	}
	return 0;
}

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