hdu 4800 Josephina and RPG dp

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

40.50 0.50 0.20 0.300.50 0.50 0.90 0.400.80 0.10 0.50 0.600.70 0.60 0.40 0.5030 1 2

 

Sample Output

0.378000

 

题意：

告诉你有m个人物，三个人可以组一个队，那么有C(m,3)种组队方法，给一个矩阵列出每个队伍之间的胜率，再告诉你有n支ai队伍以及他们的队伍编号，现在你一开始可以任选一支队伍，然后按顺序和每支ai进行比赛，你赢了ai可以选择要不要与它换队伍来进行下一场的比赛，求最大的胜率。

思路：

简单dp，d[i][j]代表第i场比赛后你的队伍编号为j时的最大胜率。

d[i][j]=max(d[i][j],mp[j][a[i]]*d[i-1][j]);
d[i][a[i]]=max(d[i][a[i]],mp[j][a[i]]*d[i-1][j]);

其中mp[i][j]为i对j的胜率，a[i]为第i支队伍的编号。

#include <iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<cmath>#define ll long long using namespace std;int n, m,a[11000];double mp[150][150],d[11000][150];int main(int argc, const char * argv[]) {    while(~scanf("%d",&m))    {        int t=m*(m-1)*(m-2)/6;        m=t;        for(int i=0;i<m;i++)            for(int j=0;j<m;j++)                scanf("%lf",&mp[i][j]);        scanf("%d",&n);        for(int i=0;i<n;i++)    scanf("%d",&a[i]);        memset(d,0,sizeof(d));        for(int i=0;i<m;i++)        {            d[0][i]=max(mp[i][a[0]],d[0][i]);            d[0][a[0]]=max(mp[i][a[0]],d[0][a[0]]);        }        for(int i=1;i<n;i++)        {            for(int j=0;j<m;j++)            {                d[i][j]=max(d[i][j],mp[j][a[i]]*d[i-1][j]);                d[i][a[i]]=max(d[i][a[i]],mp[j][a[i]]*d[i-1][j]);            }        }        double ans=0;        for(int i=0;i<m;i++)    ans=max(ans,d[n-1][i]);        printf("%.12f\n",ans);    }}