hdu3076--ssworld VS DDD(概率dp第三弹，求概率)

ssworld VS DDD

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1487    Accepted Submission(s): 304

Problem Description

One day, sssworld and DDD play games together, but there are some special rules in this games.

They both have their own HP. Each round they dice respectively and get the points P1 and P2 (1 <= P1, P2 <= 6). Small number who, whose HP to reduce 1, the same points will remain unchanged. If one of them becomes 0 HP, he loses. 

As a result of technical differences between the two, each person has different probability of throwing 1, 2, 3, 4, 5, 6. So we couldn’t predict who the final winner. 

 

Input

There are multiple test cases.

For each case, the first line are two integer HP1, HP2 (1 <= HP1, HP2 <= 2000), said the first player sssworld’s HP and the second player DDD’s HP. 

The next two lines each have six floating-point numbers per line. The jth number on the ith line means the the probability of the ith player gets point j. The input data ensures that the game always has an end. 

 

Output

One float with six digits after point, indicate the probability sssworld won the game.

 

Sample Input

5 5
1.000 0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000 1.000
5 5
0.000 0.000 0.000 0.000 0.000 1.000
1.000 0.000 0.000 0.000 0.000 0.000

 

Sample Output

0.000000
1.000000

 

Source

2009 Multi-University Training Contest 17 - Host by NUDT

求概率。和求期望的方法同样，只是不用再+1，dp[i][j]表示a有i血量，b有j血量时a赢的概率，由于要求a赢的概率。所以在dp[i][0]a赢得概率是1，dp[0][j]时a赢的概率是0。

一个坑点。血量是倒着输入的。

。。。坑了一天。。。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int hp1 , hp2 ;
double dp[2][2100] ;
double a , b , p ;
double ka[10] , kb[10] ;
int main()
{
    int i , j , flag ;
    while(scanf("%d %d", &hp2, &hp1)!=EOF)
    {
        for(i = 1 ; i <= 6 ; i++)
            scanf("%lf", &ka[i]);
        for(j = 1 ; j <= 6 ; j++)
            scanf("%lf", &kb[j]);
        memset(dp,0,sizeof(dp));
        a = b = p = 0.0 ;
        for(i = 1 ; i <= 6 ; i++)
            for(j = 1 ; j <= 6 ; j++)
            {
                if(i > j)
                    a += ka[i]*kb[j] ;
                else if( i < j )
                    b += ka[i]*kb[j] ;
                else
                    p += ka[i]*kb[j] ;
            }
        dp[0][0] = dp[1][0] = 1.0 ;
        flag = 0 ;
        for(i = 1 ; i <= hp1 ; i++)
        {
            flag = 1 - flag ;
            for(j = 0 ; j <= hp2 ; j++)
            {
                if( j == 0 ) continue ;
                dp[flag][j] = ( a*dp[flag][j-1] + b*dp[1-flag][j] ) / (1.0-p) ;
            }
        }
        printf("%.6lf\n", dp[flag][hp2]);
    }
    return 0;
}