CF# 301 D Bad Luck Island（概率dp+记忆化）_d. bad luck island-优快云博客

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探讨在BadLuck岛上，岩石、剪刀、布三种生物相遇时的生存规则及长期演化后的单一物种存活概率。

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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

题意：r吃s,s吃p,p吃r,给出三个数数量，求最后三种分别存活的概率

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 105

int r,s,p;
double dp[N][N][N];

double dfs(int pre,int now,int nest)     //pre吃now，now吃nest，nest吃pre，最后pre存活的概率
{
     if(dp[pre][now][nest]>=0) return dp[pre][now][nest];
     if(pre==0) return 0;

     if(nest==0) return pre ? 1:0;
     double ans=0;
     if(pre&&now) ans+=(pre*now*1.0)/(pre*now+now*nest+pre*nest)*dfs(pre,now-1,nest); //pre吃now
     if(now&&nest) ans+=(now*nest*1.0)/(pre*now+now*nest+pre*nest)*dfs(pre,now,nest-1); //now吃nest，
     if(pre&&nest) ans+=(pre*nest*1.0)/(pre*now+now*nest+pre*nest)*dfs(pre-1,now,nest); //nest吃pre
     dp[pre][now][nest]=ans;
     return ans;
}

int main()
{
	int i,j,n;
	mem(dp,-1);
	while(~scanf("%d%d%d",&r,&s,&p))
	{
		printf("%.10lf\n",dfs(r,s,p));
		printf("%.10lf\n",dfs(s,p,r));
		printf("%.10lf\n",dfs(p,r,s));
	}
   return 0;
}