hdu 4405 Aeroplane chess

最新推荐文章于 2020-03-28 21:53:02 发布

Jazz_Charles

最新推荐文章于 2020-03-28 21:53:02 发布

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分类专栏： dp 文章标签： dp 期望概率

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本文介绍了一种使用动态规划方法解决飞行棋游戏中达到终点的期望掷骰次数的问题。通过逆向思考并建立状态转移方程，文章提供了一个有效的算法实现。

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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

Sample Output

1.1667
2.3441

题目大意：求飞行棋从0到n的期望掷骰子点数，其中还有xi可以直接飞到yi的点，可以连飞。

思路：虽然不是很难，第一道自己写的概率dp还是应该纪念一下。求期望用逆推的方式，dp[i]表示从点i到n还需要的期望。

那么独立的考虑当前点i，要么可以直接飞走（期望直接等于目的地的期望），要么等概率1/6地去前6个点。转移方程为：

dp[i]=1/6(dp[i+1]+1) + 1/6(dp[i+2]+1) + ..... + 1/6( dp[i+6]+1)

#include <iostream>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MP(X,Y) make_pair(X,Y)
#define PB(X) push_back(X)
#define REP(X,N) for(int X=0;X<N;X++)
#define REP2(X,L,R) for(int X=L;X<=R;X++)
#define DEP(X,R,L) for(int X=R;X>L;X--)
#define DEP2(X,R,L) for(int X=R;X>=L;X--)
#define CLR(A,X) memset(A,X,sizeof(A))
#define IT iterator
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef vector<PII> VII;
typedef vector<int> VI;
#define X first
#define Y second
#define lson(X) ((X)<<1)
#define rson(X) ((X)<<1|1)
const int maxn=1e5+100;
map<int,int> mp;
double dp[maxn]; 
//dp[i]:从i到终点还需要的点数期望 
int main(){
	int n,m;
	while(~scanf("%d%d",&n,&m),n||m)
	{
		mp.clear();
		REP(i,m){
			int x,y;
			scanf("%d%d",&x,&y);
			mp[x]=y;
		}
		CLR(dp,0);
		dp[n]=0;
		for(int i=n-1;i>=0;i--){
			if(mp[i]) dp[i]=dp[mp[i]];
			else{
				dp[i]=(1.0/6)*(dp[i+1]+dp[i+2]+dp[i+3]+dp[i+4]+dp[i+5]+dp[i+6]+6);
			}
		} 
		printf("%.4f\n",dp[0]);
	}
	return 0;
}
/*
2 0
8 3
2 4
4 5
7 8
0 0
*/