概率dp

此题的巧妙之处在于倒推，如果正推状态转移无法正确进行。令f[i]表示从i到游戏结束所需要的期望步数，则（1）f[i] = f[j]，强制转移（2）f[i] = (f[j] + 1) / 6，共6种合法情况。则答案变成f[0]的值。代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
#include <stack>
#include <string>
#include <ctime>
#include <queue>
#define mem0(a) memset(a, 0, sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define eps 0.0000001
#define lowbit(x) ((x) & -(x))
#define memc(a, b) memcpy(a, b, sizeof(b))
#define xx(a) ((a) * (a))
using namespace std;
#define LL __int64
#define DB double
#define pi 3.14159265359
#define MD 10000007
#define INF (int)1e9
double f[110000];
int last[110000];
int main()
{
        //freopen("input.txt", "r", stdin);
        int n, m;
        while(cin>> n>> m, n || m) {
                mem0(last);
                for(int i = 1; i <= m; i++) {
                        int u, v;
                        cin>> u>> v;
                        last[u] = v;
                }
                mem0(f);
                for(int i = n - 1; i >= 0; i--) {
                        if(last[i]) {
                                f[i] = f[last[i]];
                                continue;
                        }
                        for(int j = 1; j <= 6; j++) f[i] += (f[i + j] + 1) / 6;
                }
                printf("%.4f
", f[0]);
        }
        return 0;
}