HDU 4405 Aeroplane chess 期望dp

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4405

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
#### 问题描述
> 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
> Please help Hzz calculate the expected dice throwing times to finish the game.
#### 输入
> There are multiple test cases.
> Each test case contains several lines.
> The first line contains two integers N(1≤N≤100000) and M(0≤M≤1000).
> Then M lines follow, each line contains two integers Xi,Yi(1≤Xi The input end with N=0, M=0.

输出

For each test case in the input, you should output a line indicating the expected dice throwing times. Output should be rounded to 4 digits after decimal point.

样例输入

2 0

8 3

2 4

4 5

7 8

0 0

样例输出

1.1667

2.3441

题意

飞行棋，每次投掷骰子，(1-6等概率),按骰子大小前进，其中有若干个飞机场(a,b)，可以从a直接传送到b，现从0点出发，问到达>=N的点需要的期望步数。

题解

期望dp入门：

全期望公式：E[x]=sigma(pi*E[xi]).

有飞机场的话：E[a]=E[b]。

否则：E[i]=sigma(1/6*(E[i+x]+1)).

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=101010;

///dp[i]表示在i点，到达终点还需几步
///期望一般逆推求，比如这一题，你已知的状态是终点而不是起点。
double dp[maxn];
int mp[maxn];
int n,m;

void init(){
    clr(mp,-1);
    clr(dp,0);
}

int main() {
    while(scf("%d%d",&n,&m)==2&&n){
        init();
        rep(i,0,m){
            int u,v;
            scf("%d%d",&u,&v);
            mp[u]=v;
        }
        for(int i=n-1;i>=0;i--){
            ///有飞机场，直接飞过去
            if(mp[i]!=-1){
                dp[i]=dp[mp[i]];
                continue;
            }
            ///掷骰子,全期望公式
            for(int x=1;x<=6;x++){
                dp[i]+=1.0/6*(dp[i+x]+1);
            }
        }
        prf("%.4lf\n",dp[0]);
    }
    return 0;
}

//end-----------------------------------------------------------------------