zoj 3329 One Person Game 概率DP
2013-08-07 21:02
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思路:这题的递推方程有点麻烦!!
dp[i]表示分数为i的期望步数,p[k]表示得分为k的概率,p0表示回到0的概率:
dp[i]=Σ(p[k]*dp[i+k])+dp[0]*p0+1
设dp[i]=A[i]*dp[0]+B[i]带入的:
dp[i]=∑(pk*A[i+k]*dp[0]+pk*B[i+k])+dp[0]*p0+1
=(∑(pk*A[i+k])+p0)dp[0]+∑(pk*B[i+k])+1;
明显A[i]=(∑(pk*A[i+k])+p0)
B[i]=∑(pk*B[i+k])+1
先递推求得A[0]和B[0].
那么 dp[0]=B[0]/(1-A[0]);
链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3754
代码如下:
View Code
dp[i]表示分数为i的期望步数,p[k]表示得分为k的概率,p0表示回到0的概率:
dp[i]=Σ(p[k]*dp[i+k])+dp[0]*p0+1
设dp[i]=A[i]*dp[0]+B[i]带入的:
dp[i]=∑(pk*A[i+k]*dp[0]+pk*B[i+k])+dp[0]*p0+1
=(∑(pk*A[i+k])+p0)dp[0]+∑(pk*B[i+k])+1;
明显A[i]=(∑(pk*A[i+k])+p0)
B[i]=∑(pk*B[i+k])+1
先递推求得A[0]和B[0].
那么 dp[0]=B[0]/(1-A[0]);
链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3754
代码如下:
#include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<cstring> #include<vector> #define ll __int64 #define pi acos(-1.0) #define MAX 600 using namespace std; double p[100],A[MAX],B[MAX],p0; int main(){ int n,i,j,t,k,s,k1,k2,k3,a,b,c; cin>>t; while(t--){ cin>>n>>k1>>k2>>k3>>a>>b>>c; p0=1.0/k1/k2/k3; memset(p,0,sizeof(p)); for(i=1;i<=k1;i++) for(j=1;j<=k2;j++) for(k=1;k<=k3;k++){ if(i!=a||j!=b||k!=c) p[i+j+k]+=p0; } memset(A,0,sizeof(A)); memset(B,0,sizeof(B)); for(i=n;i>=0;i--){ A[i]=p0;B[i]=1; for(j=1;j<=k1+k2+k3;j++){ A[i]+=A[i+j]*p[j]; B[i]+=B[i+j]*p[j]; } } printf("%.15lf\n",B[0]/(1.0-A[0])); } return 0; }
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