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

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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;}

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