codeforces 540D Bad Luck Island (概率DP)

题意：会出石头、剪刀、布的人分别有r,s,p个，他们相互碰到的概率相同，输的人死掉，问最终活下去的人是三种类型的概率

设状态dp(i,j,k)为还有i个石头，j个剪刀，k个布时的概率，dp(r,s,p)=1.0

状态转移方程:

d(i-1,j,k)+=d(i,j,k)*(i*k)/(i*j+i*k+j*k);

d(i,j-1,k)+=d(i,j,k)*(i*j)/(i*j+i*k+j*k);

d(i,j,k-1)+=d(i,j,k)*(j*k)/(i*j+i*k+j*k);

因为状态dp(i,j,k)可以由dp(i+1,j,k)、dp(i,j+1,k)和dp(i,j,k+1)转移过来，所以用+=

一个三重循环解决，复杂度O(n^3)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
    
#define fst first
#define sc second
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;


const int maxn = 5e5 + 100;
const int maxm = 5e5 + 100;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-6;
inline int read(){
    
    int num;
    char ch;
    while((ch=getchar())<'0' || ch>'9');
    num=ch-'0';
    while((ch=getchar())>='0' && ch<='9'){
        num=num*10+ch-'0';
    }
    return num;
}
double dp[101][101][101];
int main(){
    int r, s ,p;
    scanf("%d %d %d", &r, &s, &p);
    mem(dp, 0);
    dp[r][s][p] = 1;
    for(int i = r; i >= 1; i--){
        for(int j = s; j >= 1; j--){
            for(int k = p; k >= 1; k--){
                double sum = i*j + j*k + i*k;
                printf("%d %d %d %lf
", i, j, k, dp[i][j][k]);
                dp[i-1][j][k] = dp[i][j][k]*(double)(i*k)/sum;
                dp[i][j-1][k] = dp[i][j][k]*(double)(i*j)/sum;
                dp[i][j][k-1] = dp[i][j][k]*(double)(j*k)/sum;
            }
        }
    }
    double ans1, ans2, ans3;
    ans1 = ans2 = ans3 = 0;
    for(int i = 1; i <= 100; i++){
        for(int j = 1; j <= 100; j++){
            ans1 += dp[i][j][0];
            ans2 += dp[0][i][j];
            ans3 += dp[j][0][i];
        }
    }
    printf("%.17lf %.17lf %.17lf", ans1, ans2, ans3);
    return 0;
}
/*

*/