2017寒假概率dp训练题

poj 3744 Scout YYF I    

简单题,不难找到递推,设dp[i]为安全达到i的概率,dp[0]=0,dp[1]=1,dp[i]=p*dp[i-1]+(1-p)*dp[i-2],用矩阵快速幂计算,再相乘.

代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int N;double p;
int a[15];
struct matrix{
    double a[2][2];
};
matrix muti(matrix x,matrix y){
    matrix t;
    for(int i=0;i<2;++i){
        for(int j=0;j<2;++j){
            t.a[i][j]=0;//init
            for(int k=0;k<2;++k){
                t.a[i][j]+=1.0*x.a[i][k]*y.a[k][j];
            }
        }
    }
    return t;
}
double getP(int n){
    matrix ans,base;
    ans.a[0][0]=1;ans.a[0][1]=ans.a[1][0]=ans.a[1][1]=0;
    base.a[0][0]=p;base.a[0][1]=1-p;base.a[1][0]=1.0;base.a[1][1]=0.0;
    if(n<0)return 0;
    while(n){
    //    printf("n=%d
",n);
        if(n&1)ans=muti(base,ans);
        base=muti(base,base);
        n/=2;
    }
    return ans.a[0][0];
}
int main(){
    while(~scanf("%d%lf",&N,&p)){
        for(int i=0;i<N;++i){
            scanf("%d",&a[i]);
        }
        sort(a,a+N);
        double res=1.0;
        int prev=1;
        for(int i=0;i<N;++i){
            res=res*(1-p)*getP(a[i]-prev-1);
            prev=a[i]+1;
        }
        printf("%.7f
",res);
    }
    return 0;
}

 cf148D

题目链接:http://codeforces.com/problemset/problem/148/D

简单题,我是写的记忆化搜索,但开始在巨龙的公式那出了问题,花了两个多小时.

代码:

#include<cstdio>
#include<cstring>
const int MAXN=1007;
double dp[MAXN][MAXN][2];
int visit[MAXN][MAXN][2];
//0 is princess,1 is dragon
double dfs(int w,int b,int status){
    //printf("w=%d,b=%d,status=%d
",w,b,status);
    if(w<=0){
         dp[0][b][1]=1.0;
         dp[0][b][0]=0.0;
         return dp[0][b][status];
    }
    if(b<=0){
        dp[w][0][status]=1.0;
        dp[w][0][1-status]=0.0;
        return dp[w][0][status];
    }
    if(visit[w][b][status])return dp[w][b][status];
    visit[w][b][status]=1;
    double res=w*1.0/(w+b);
    if(status==0){
        res+=b*1.0/(w+b)* (1.0-dfs(w,b-1,1));
    }else{
        res+=b*1.0/(w+b)* (((w*1.0/(w+b-1)*(1-dfs(w-1,b-1,0)))) + ((b-1)*1.0/(w+b-1)*(1-dfs(w,b-2,0))));
    }
    dp[w][b][status]=res;
    //printf("dp[%d][%d][%d]=%.9f
",w,b,status,res);
    return dp[w][b][status];
}
int main(){
    int w,b;
    scanf("%d%d",&w,&b);
    memset(visit,0,sizeof(visit));
    printf("%.10f
",dfs(w,b,0));
    return 0;
}

 hdoj4405 飞行棋

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4405

期望入门题,问达到终点期望投多少次骰子.

代码:

#include<cstdio>
#include<cstring>
const int MAXN=2e5+7,MAXM=1e3+7;
double dp[MAXN];
int fly[MAXN];
int main(){
    int N,M;
    while(scanf("%d%d",&N,&M)&&N+M){
        memset(fly,0,sizeof(fly));
        memset(dp,0,sizeof(dp));
        for(int i=0;i<M;++i){
            int a,b;scanf("%d%d",&a,&b);
            fly[a]=b;
        }
        dp[N]=0;
        for(int i=N-1;i>=0;--i){
            if(fly[i]==0){
                for(int x=1;x<=6;++x){
                    dp[i]+=1.0/6*dp[i+x];
                }
                dp[i]++;
            }
            else dp[i]=dp[fly[i]];
        }
        printf("%.4f
",dp[0]);
    }
    return 0;
}