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:
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 int N;double p; 6 int a[15]; 7 struct matrix{ 8 double a[2][2]; 9 }; 10 matrix muti(matrix x,matrix y){ 11 matrix t; 12 for(int i=0;i<2;++i){ 13 for(int j=0;j<2;++j){ 14 t.a[i][j]=0;//init 15 for(int k=0;k<2;++k){ 16 t.a[i][j]+=1.0*x.a[i][k]*y.a[k][j]; 17 } 18 } 19 } 20 return t; 21 } 22 double getP(int n){ 23 matrix ans,base; 24 ans.a[0][0]=1;ans.a[0][1]=ans.a[1][0]=ans.a[1][1]=0; 25 base.a[0][0]=p;base.a[0][1]=1-p;base.a[1][0]=1.0;base.a[1][1]=0.0; 26 if(n<0)return 0; 27 while(n){ 28 // printf("n=%d\n",n); 29 if(n&1)ans=muti(base,ans); 30 base=muti(base,base); 31 n/=2; 32 } 33 return ans.a[0][0]; 34 } 35 int main(){ 36 while(~scanf("%d%lf",&N,&p)){ 37 for(int i=0;i<N;++i){ 38 scanf("%d",&a[i]); 39 } 40 sort(a,a+N); 41 double res=1.0; 42 int prev=1; 43 for(int i=0;i<N;++i){ 44 res=res*(1-p)*getP(a[i]-prev-1); 45 prev=a[i]+1; 46 } 47 printf("%.7f\n",res); 48 } 49 return 0; 50 }
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:
1 #include<cstdio> 2 #include<cstring> 3 const int MAXN=1007; 4 double dp[MAXN][MAXN][2]; 5 int visit[MAXN][MAXN][2]; 6 //0 is princess,1 is dragon 7 double dfs(int w,int b,int status){ 8 //printf("w=%d,b=%d,status=%d\n",w,b,status); 9 if(w<=0){ 10 dp[0][b][1]=1.0; 11 dp[0][b][0]=0.0; 12 return dp[0][b][status]; 13 } 14 if(b<=0){ 15 dp[w][0][status]=1.0; 16 dp[w][0][1-status]=0.0; 17 return dp[w][0][status]; 18 } 19 if(visit[w][b][status])return dp[w][b][status]; 20 visit[w][b][status]=1; 21 double res=w*1.0/(w+b); 22 if(status==0){ 23 res+=b*1.0/(w+b)* (1.0-dfs(w,b-1,1)); 24 }else{ 25 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)))); 26 } 27 dp[w][b][status]=res; 28 //printf("dp[%d][%d][%d]=%.9f\n",w,b,status,res); 29 return dp[w][b][status]; 30 } 31 int main(){ 32 int w,b; 33 scanf("%d%d",&w,&b); 34 memset(visit,0,sizeof(visit)); 35 printf("%.10f\n",dfs(w,b,0)); 36 return 0; 37 }
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:
1 #include<cstdio> 2 #include<cstring> 3 const int MAXN=2e5+7,MAXM=1e3+7; 4 double dp[MAXN]; 5 int fly[MAXN]; 6 int main(){ 7 int N,M; 8 while(scanf("%d%d",&N,&M)&&N+M){ 9 memset(fly,0,sizeof(fly)); 10 memset(dp,0,sizeof(dp)); 11 for(int i=0;i<M;++i){ 12 int a,b;scanf("%d%d",&a,&b); 13 fly[a]=b; 14 } 15 dp[N]=0; 16 for(int i=N-1;i>=0;--i){ 17 if(fly[i]==0){ 18 for(int x=1;x<=6;++x){ 19 dp[i]+=1.0/6*dp[i+x]; 20 } 21 dp[i]++; 22 } 23 else dp[i]=dp[fly[i]]; 24 } 25 printf("%.4f\n",dp[0]); 26 } 27 return 0; 28 }
