HDU-4405 Aeroplane chess 期望DP

dp[i]表示第i个位置跳出去的期望天数，先构造出N+1到N+5这几个位置，然后先把dp[N-N+5]这六个位置全部赋值为0，因为这几个位置都已经出去了。

然后就是递推了

如果该点没有航班的话：
dp[x] = (1/6)*(dp[x+1] + dp[x+2] + dp[x+3] + dp[x+4] + dp[x+5] + dp[x+6]) + 1;

否则：
dp[x] = dp[link[x]]; 其中link[x]表示x连到哪一个点。

代码如下：

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<algorithm>

#include<cmath>

#include<queue>

#include<set>

#include<map>

#include<cstring>

#include<vector>

#include<string>

#define MAXN 100020

#define LL long long

using namespace std;



int N, M;



const double EE = 1.0/6;



double dp[MAXN];



int link[MAXN];



int main(  )

{

    int len, a, b;

    while (scanf("%d %d", &N, &M) == 2) {

        len = N+5;

        memset(link, 0xff, sizeof (link));

        for (int i = N; i <= len; ++i) dp[i] = 0;

        for (int i = 1; i <= M; ++i) {

            scanf("%d %d", &a, &b);

            link[a] = b;

        }

        for (int i = N-1; i >= 0; --i) {

            double temp = 0.0;

            if (link[i] != -1) {

                temp = dp[link[i]];

            } else {

                for (int j = i + 1; j <= i + 6; ++j) {

                    temp += EE * dp[j];

                }

                temp += 1;

            }

            dp[i] = temp;

        }

        printf("%.4lf\n", dp[0]);

    }

    return 0;

}