【期望DP】HDU4405 Aeroplane chess

题面在这里

还是期望DP的老套路……

从后往前DP，fi表示i到终点的期望值，然后XJB转移就好了

还有题意貌似是能飞就必须飞？没讲清楚啊

示例程序：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=100010;int n,m,to[maxn];double f[maxn];int main(){    scanf("%d%d",&n,&m);    while (n){        for (int i=0;i<=n;i++) to[i]=i;        for (int i=1,x,y;i<=m;i++) scanf("%d%d",&x,&y),to[x]=y;        memset(f,0,sizeof(f));        for (int i=n-1;i>=0;i--){            if (to[i]>i) {f[i]=f[to[i]];continue;}            f[i]=1;            for (int j=1;j<=6;j++)             f[i]+=f[i+j]/6;        }        printf("%.4lf\n",f[0]);        scanf("%d%d",&n,&m);    }    return 0;}