#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <algorithm>#include <vector>#include <cmath>#include <queue>#include <set>#define LL long long#define FOR(i,x,y) for(int i=x;i<=y;i++)using namespace std;const int maxn = 100000 + 10;double dp[maxn];vector<int> vec[maxn];int N , M;int main(){while(scanf("%d%d", &N, &M)!=EOF){if(N == 0 && M == 0)break;FOR(i,1,N) vec[i].clear();FOR(i,1,M){int x, y;scanf("%d %d", &x, &y);vec[x].push_back(y);}memset(dp, 0, sizeof(dp));dp[N] = 0;for(int i=N-1;i>=0;i–){if(vec[i].size() > 0){dp[i] = dp[vec[i][0]];}else{for(int x=1;x<=6;x++){dp[i] += (1.0 / 6.0) * dp[i+x];}dp[i] += 1;}}printf("%.4lf\n", dp[0]);}return 0;}题目大意：
Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces are 1,2,3,4,5,6). When Hzz is at grid i and the dice number is x, he will moves to grid i+x. Hzz finishes the game when i+x is equal to or greater than N.There are also M flight lines on the chess map. The i-th flight line can help Hzz fly from grid Xi to Yi (0<Xi<Yi<=N) without throwing the dice. If there is another flight line from Yi, Hzz can take the flight line continuously. It is granted that there is no two or more flight lines start from the same grid.Please help Hzz calculate the expected dice throwing times to finish the game.
解题思路：
求概率，依旧是用逆推的方式求解。
dp[i] 表示在 点 i 处走到终点所需要掷骰子的期望次数。
很容易写出递推方程。
当点 i 处可以跳跃到点 j 时 dp[i] = dp[j];
当点 i 处不可以跳跃时， dp[i] = sigma(dp[i+x] / 6)
，征服畏惧、建立自信的最快最确实的方法，
