HDU4800——Josephina and RPG(概率dp)

Josephina and RPG

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1538    Accepted Submission(s): 467
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

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

 

题意不说了，自己读吧= =

dp[i][j]表示打完前i个人，自己保留的队伍为j的概率。

a[i]表示第i个AI指向的队伍。那么如果j不等于a[i]，那么意味着打完第i-1个人就得是j号AI，如果J=a[i]，那么打完第i-1个人可以是任意一个AI，这样就能得到递推关系了。

#include <cmath>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <algorithm>#include <string>#include <map>#include <queue>#include <set>#include <stack>#include <ctime>#include <cstdlib>#include <iostream>#define pi 3.1415926#define mod 1000000007#define MAXN 200#define INF 0x3f3f3f3f3f3f3fLLusing namespace std;double f[MAXN][MAXN];int a[10010];double dp[10010][MAXN];int main(){int x;while(scanf("%d",&x)!=EOF){int n=x*(x-1)*(x-2)/6;for(int i=0;i<n;i++){for(int j=0;j<n;j++){scanf("%lf",&f[i][j]);}}int m;scanf("%d",&m);for(int i=0;i<m;i++){for(int j=0;j<n;j++)dp[i][j]=0;}for(int i=0;i<m;i++)scanf("%d",a+i);double ans=0;for(int i=0;i<n;i++){dp[0][i]=max(dp[0][i],f[i][a[0]]);ans=max(dp[0][i],ans);}dp[0][a[0]]=ans;ans=0;for(int i=1;i<m;i++){for(int j=0;j<n;j++){if(j!=a[i])dp[i][j]=max(dp[i][j],f[j][a[i]]*dp[i-1][j]);//ans=max(dp[i][j],ans);}for(int j=0;j<n;j++){dp[i][a[i]]=max(dp[i][a[i]],dp[i-1][j]*f[j][a[i]]);}}ans=0;for(int j=0;j<n;j++){ans=max(ans,dp[m-1][j]);}// for(int i=0;i<m;i++){// for(int j=0;j<n;j++){// printf("%f ",dp[i][j]);// }// printf("\n");// }printf("%.6lf\n",ans);}}