hdoj4405-Aeroplane chess

求期望值，体重说明，如果可以flight的话就直接到达规定那个点，用f数组将两个点连在一起就可以，反之，就要求打骰子后的期望啦，e[i] =∑(e[i + k] * 1 / 6) + 1（k为点数1~6,1/6为致筛子的概率，因为不可以flight就要置一次筛子，所以+1）。

/************************************************************************* > File Name: hdoj4405.cpp > Author: AcToy > Mail: [e-mail] > Created Time: 2013年07月17日 星期三 09时19分05秒 ************************************************************************/#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <climits>#include <cstdio>#include <string>#include <vector>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>using namespace std;typedef unsigned int u32;typedef long long i64;typedef unsigned long long u64;typedef vector<int> IV;typedef vector<bool> BV;typedef pair<int,int> II;typedef vector<II> IIV;#define For(t,v,c) for(t::const_iterator v=c.begin(); v!=c.end(); ++v)const int INF = 0x7FFFFFFF;const double eps = 1E-10;const double PI = acos(-1);const int maxn = 100000;double e[maxn];int f[maxn], n, m;int main () {    while(scanf("%d%d", &n, &m) == 2 && n) {        memset(f, 0, sizeof(f));        for(int i = 0; i <= n; e[i] = 0, ++i) ;        for(int i = 1; i <= m; ++i) {            int x, y;            scanf("%d%d", &x, &y);            f[x] = y;        }        for(int i = n - 1; i >= 0; --i) {            if(f[i]) e[i] = e[f[i]];            else {                for(int j = 1; j <= 6; ++j) {                    if(i + j <= n) e[i] += e[i + j] * 1.0 / 6;                    else break;                }                e[i] += 1.0;            }        }        printf("%.4lf\n", e[0]);    }    return 0;}