Josephina and RPG HDU - 4800
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.
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
題目大意就是: RPG遊戲中有很多不同的角色,選擇其中三個組成一支隊伍,現已知每支隊伍打敗其他隊伍的概率,然後給你一個需要打敗的隊伍的順序,求打敗所有隊伍的最大的概率是多少。
其實就是一道概率DP的題目,換或者不換。
對需要打敗的隊伍進行枚舉,考慮換了隊伍和不換隊伍兩者概率大小,取最大的那種。
// Asimple #include <iostream> #include <algorithm> #include <cstdio> #include <cstdlib> #include <string> #include <cstring> #include <stack> #include <set> #include <map> #include <cmath> #define INF 0x3f3f3f3f #define debug(a) cout<<#a<<" = "<<a<<endl #define test() cout<<"============"<<endl #define CLS(a,v) memset(a, v, sizeof(a)) #define pas system("pause") using namespace std; typedef long long ll; typedef unsigned long long ull; int dx[] = {-1,1,0,0,-1,-1,1,1}, dy[]={0,0,-1,1,-1,1,1,-1}; const int maxn = 10000+5; const ll mod = 1000000007; ll n, m, T, len, cnt, num, ans, Max, k; //vector<int> a[maxn]; double dp[maxn]; double Map[400][400]; int team[maxn]; void input(){ while( scanf("%lld", &n)!=EOF) { n = n*(n-1)*(n-2)/6; double res = 1.0; for(int i=0; i<n; i++) { for(int j=0; j<n; j++) { scanf("%lf", &Map[i][j]); } dp[i] = 1.0; } scanf("%lld", &m); for(int i=0; i<m; i++) scanf("%d", &team[i]); for(int i=0; i<m; i++) { double te = 0.0; for(int j=0; j<n; j++) { te = max(Map[j][team[i]]*dp[j], te); dp[j] = dp[j] * Map[j][team[i]]; } dp[team[i]] = te; } printf("%.6lf\n", dp[team[m-1]]); } // pas; } int main() { input(); return 0; }
Josephina and RPG HDU - 4800