LightOJ-1395 A Dangerous Maze (II) (期望dp)

P - A Dangerous Maze (II)

 LightOJ - 1395 

题解：期望dp，设正确的门有a个，平均耗时为sum1，错误的门有b个，平均耗时为sum2。

状态转移分2部分：①当i==k时，即已经记住k次走错的门，dp[i]=a*sum1/(n-k)+(dp[k]+sum2)*(b-k)/(n-k)

②当i<k时，dp[i]=a/(n-i)*sum1+(b-i)/(n-i)*(dp[i+1]+sum2)

#include<bits/stdc++.h>
using namespace std;
const int MX = 105;
double dp[MX];
int x[MX];
int main(){
int T,n,k;
// freopen("in.txt","r",stdin);
scanf("%d",&T);
for(int cas=1;cas<=T;cas++){
memset(dp,0,sizeof(dp));
scanf("%d%d",&n,&k);
int a=0,b;
double sum1=0,sum2=0;
for(int i=0;i<n;i++) {
scanf("%d",&x[i]);
if(x[i]>0) {sum1+=x[i];a++;}
else sum2+=abs(x[i]);
}
if(a==0) {
printf("Case %d: -1\n",cas);
continue;
}
b=n-a;
if(a) sum1/=a;
if(b) sum2/=b;
k=min(k,b);
dp[k]=sum1+(b-k)*sum2/a;
for(int i=k-1;i>=0;i--){
dp[i]=(dp[i+1]+sum2)*(b-i);
dp[i]+=sum1*a;
dp[i]/=n-i;
}
printf("Case %d: %.7f\n",cas,dp[0]);
}
return 0;
}