HDU4405-Aeroplane chess（可能性DP需求预期）

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1182    Accepted Submission(s): 802

Problem Description

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.

 

Input

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<Yi≤N).  
The input end with N=0, M=0. 

 

Output

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.

 

Sample Input

   
     

      
    2 0
8 3
2 4
4 5
7 8
0 0
   
     

 

Sample Output

   
     

      
    1.1667
2.3441
   
     

 

题意:从0点出发，掷筛子。筛子掷到几（概率同样）就走几步。当中还有像飞行棋一样的设置，即能够从某一点直接跳到下一点。并且还能够连续跳跃，问你从0点到n的步数的数学期望。

思路：dp[i] 表示眼下走到第i格。距离走到终点的数学期望。递推就可以

前期用dfs重构一下图就能够了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;
const int maxn = 100000+10;
double dp[maxn];
map<int,int> t;
map<int,int> flight;
int n,m;
void dfs(int sta,int x){
    if(t[x]==0){
        flight[sta] = x;
    }else{
        dfs(sta,t[x]);
    }
}
void init(){
    for(map<int,int>::iterator it = t.begin(); it != t.end(); it++){
        dfs(it->first,it->second);
    }
}
int main(){

    while(cin >> n >> m && n+m){
        t.clear();
        flight.clear();
        while(m--){
            int a,b;
            scanf("%d%d",&a,&b);
            t[a] = b;
        }
        init();
        dp[n] = 0;
        for(int i = n-1; i >= 0; i--){
            dp[i] = 1;
            for(int k = 1; k <= 6; k++){
                if(i+k <= n){
                    if(flight[i+k]==0){
                        dp[i] += dp[i+k]/6;
                    }else{
                        dp[i] += dp[flight[i+k]]/6;
                    }

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

    }
    return 0;
}

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