Codeforces Round #301 (Div. 2) -- D. Bad Luck Island （概率DP）

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The Bad Luck Island is inhabited by three kinds of species: r rocks, s scissors and p papers. At some moments of time two random individuals meet (all pairs of individuals can meet equiprobably), and if they belong to different species, then one individual kills the other one: a rock kills scissors, scissors kill paper, and paper kills a rock. Your task is to determine for each species what is the probability that this species will be the only one to inhabit this island after a long enough period of time.

Input

The single line contains three integers r, s and p (1 ≤ r, s, p ≤ 100) — the original number of individuals in the species of rock, scissors and paper, respectively.

Output

Print three space-separated real numbers: the probabilities, at which the rocks, the scissors and the paper will be the only surviving species, respectively. The answer will be considered correct if the relative or absolute error of each number doesn't exceed 10 - 9.

Sample test(s)
input
2 2 2
output
0.333333333333 0.333333333333 0.333333333333
input
2 1 2
output
0.150000000000 0.300000000000 0.550000000000
input
1 1 3
output
0.057142857143 0.657142857143 0.285714285714

链接：http://codeforces.com/contest/540/problem/D

大体题意：

岛上有三个物种，石头r个，剪刀s个，布p个，冤家路窄，每天都会相遇两个物种，其中石头能杀掉剪刀，布能杀掉石头，剪刀能杀掉布，问最后只剩下石头，剪刀，布的三个概率是多少？

思路：

很明显是概率DP，比赛果然没有什么思路 = =：

令dp[i][j][k]表示剩下i个石头，剩下j 个剪刀的，剩下k 个布的概率！

转移：

三个情况嘛，那一天要么石头被吃，要么剪刀被吃，要么布被吃：

石头被吃： dp[i-1][j][k] = dp[i][j][k] * i*k/sum; 就是所有的布遇到所有的石头是事件总和，也就是i*k,而sum 是所所有的情况 sum = i*j + i*k + j*k;

其余的类似：

剪刀被吃：dp[i][j-1][k] += dp[i][j][k] * i*j*1.0/sum;

布被吃：dp[i][j][k-1] += dp[i][j][k] * j*k*1.0/sum;

最后统计剩下那个物种累计求和即可！

详细见代码：

#include<bits/stdc++.h>using namespace std;const int maxn = 100 + 10;double dp[maxn][maxn][maxn];int main(){    int r,s,p;    scanf("%d %d %d",&r,&s,&p);    dp[r][s][p] = 1.0;    for (int i = r; i >= 0; --i){        for (int j = s; j >= 0; --j){            for (int k = p; k >= 0; --k){                if (i == 0 && j == 0 && k == 0)goto TT;                double sum = i*j + i*k + k*j;                if (i && k)dp[i-1][j][k] += dp[i][j][k] * i*k*1.0/sum;                if (j && i)dp[i][j-1][k] += dp[i][j][k] * i*j*1.0/sum;                if (k && j)dp[i][j][k-1] += dp[i][j][k] * j*k*1.0/sum;            }        }    }    TT:    double ans1 = 0.0,ans2 = 0.0,ans3 = 0.0;    for (int i = 1; i <= 100; ++i){        ans1 += dp[i][0][0];        ans2 += dp[0][i][0];        ans3 += dp[0][0][i];    }    printf("%.14lf %.14lf %.14lf\n",ans1,ans2,ans3);    return 0;}

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