首页 > ACM题库 > HDU-杭电 > HDU 4800-Josephina and RPG-动态规划-[解题报告]HOJ
2015
09-18

HDU 4800-Josephina and RPG-动态规划-[解题报告]HOJ

题目链接:

ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5081

HDU:http://acm.hdu.edu.cn/showproblem.php?pid=4800

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

题意:

给定C(m,3)支队伍之间对战获胜的概率,再给定一个序列存放队伍的编号,每次获胜之后可以选择和当前战胜的对手换队伍或者不换。

求挑战胜利的最大概率。

PS:

dp[i][j]:战胜第i支队伍时,当前的队伍为j!

代码如下:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 147;
double dp[10017][maxn], p[maxn][maxn];
//dp[i][j]:战胜第i支队伍时,当前的队伍为j!
int a[10017];
int n, m;

void solve()
{
    for(int i = 0; i <= m; i++)
    {
        dp[0][i] = 1;//初始值
    }
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            dp[i+1][j] = max(dp[i+1][j], dp[i][j]*p[j][a[i+1]]);//不换
            dp[i+1][a[i+1]] = max(dp[i+1][a[i+1]], dp[i][j]*p[j][a[i+1]]);//换
        }
    }
}
int main()
{
    while(~scanf("%d",&m))
    {
        memset(dp,0,sizeof(dp));
        m = m*(m-1)*(m-2)/6;
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < m; j++)
            {
                scanf("%lf",&p[i][j]);
            }
        }
        scanf("%d",&n);
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&a[i]);
        }
        solve();
        double ans = 0;
        for(int i = 0; i < m; i++)
        {
            ans = max(ans,dp[n][i]);
        }
        printf("%.6lf\n",ans);
    }
    return 0;
}

参考:http://blog.csdn.net/u012860063/article/details/40457741