HDU 3076：ssworld VS DDD（概率DP）

http://acm.split.hdu.edu.cn/showproblem.php?pid=3076

ssworld VS DDD

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

题意：两个人玩游戏，给出各自掷骰子得到不同点数的概率，点数小的扣一血，相同不扣血。

思路：概率DP，处理到第二个人血量剩余1的时候，然后再处理他降到0的时候的情况，不能直接循环第二个人血量为0的情况，即ans += dp[i][0] (1 <= i <= h1)，因为这样的话dp[i][0]继承了dp[i+1][0]的状态，但是为0的时候游戏已经停止了，这样是不合理的。应该是ans += dp[i][1] * win / (1 - ave)，这样的话才不会错，要注意考虑（1-ave）的情况，因为当前回合如果平局的话，下一个回合还是有机会赢的。（PS：很感谢ZLW师兄帮我debug了这题好久233）

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
const double eps = 0.00000001;
double d[2][6];
double dp[2005][2005];

int main()
{
    int h1, h2;
    while(~scanf("%d%d", &h2, &h1)) { //题目输入反了
        double win = 0, lose = 0, ave = 0;
        for(int i = 0; i < 2; i++)
            for(int j = 0; j < 6; j++)
                scanf("%lf", &d[i][j]);
        for(int i = 0; i < 6; i++) {
            for(int j = 0; j < 6; j++) {
                if(i == j) ave += d[0][i] * d[1][j];
                else if(i < j) lose += d[0][i] * d[1][j];
                else win += d[0][i] * d[1][j];
            }
        }
//        printf("SIZE : %d
", sizeof(dp));
        double ans = 0;
        if(eps >= fabs(1 - ave)) {
            printf("%.6f
", ans);
            continue;
        }
        for(int i = 0; i <= h1+1; i++)
            for(int j = 0; j <= h2+1; j++)
                dp[i][j] = 0;
        // dp[h1+1][h2-1] = dp[h1-1][h2+1] = dp[h1+1][h2] = dp[h1][h2+1] = 0;
        dp[h1][h2] = 1;
        for(int i = h1; i >= 0; i--) {
            for(int j = h2; j >= 0; j--) {
                if(i == h1 && j == h2) continue;
                dp[i][j] = (dp[i+1][j] * lose + dp[i][j+1] * win) / ((double)1 - ave);
            }
        }
        for(int i = 1; i <= h1; i++)
            ans += dp[i][1] * win / ((double)1 - ave);
        printf("%.6f
", ans);
    }
    return 0;
}