LightOJ-1395 A Dangerous Maze (II)

期望dp

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
const int N=100+10;
double d[N];
int main()
{
    int T,kase=0;
    scanf("%d",&T);
    while(T--)
    {
        int n,k;
        double x,sum1=0.0,sum2=0.0;
        int num1=0,num2=0;
        scanf("%d%d",&n,&k);
        for(int i=0;i<n;i++)
        {
            scanf("%lf",&x);
            if(x>0.0) sum1+=x,num1++;
            else sum2+=-x,num2++;
        }
        printf("Case %d: ",++kase);
        if(num1==0)
        {
            printf("-1\n");
            continue;
        }
        if(num1) sum1/=num1;
        if(num2) sum2/=num2;
        k=min(k,num2);
        d[k]=sum1+(num2-k)*sum2/num1;
        for(int i=k-1;i>=0;i--)
            d[i]=(sum1*num1+(d[i+1]+sum2)*(num2-i))/(n-i);
        printf("%.6f\n",d[0]);
    }
    return 0;
}