[HDU] 4405 Aeroplane chess 期望dp入门

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本文介绍了一种解决飞机棋游戏中投掷骰子次数期望问题的动态规划算法。游戏在一个包含N+1个格点的数轴上进行，玩家从起点0出发，目标是到达或超过终点N。游戏过程中，玩家可以通过掷骰子前进，还能利用飞行线路快速前进。文章详细解析了如何通过逆向思维和状态转移方程计算平均投掷次数。

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4454 Accepted Submission(s): 2809

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

Source

2012 ACM/ICPC Asia Regional Jinhua Online

题目大意 : 数轴上有N+1个点（编号0~N），一个人玩游戏，从0出发，当到达N或大于N的点则游戏结束。每次行动掷骰子一次，骰子编号1-6，掷到多少就向前走几步，这个数轴上还有些特殊点，这些点类似飞行棋中的飞行点，只要到达这些点就可以直接飞到给定点。求总共投掷骰子次数的期望。

题解：入门期望dp，对于期望，我们的通常做法都是倒着来，这道题也不例外. 设dp[i]为在i位置上到终点的掷骰子需要期望的次数，那么dp[>=n]显然为0，我们要求的是dp[0]. dp[i]可以由dp[i+1], dp[i+2]...转移过来，并且对于每一个的概率都是1/6. 当然如果这个点为特殊点的话，那么dp[i] = dp[可以飞往的点].

话说第一次不看题解做出来，虽然本来这题就很简单...不过还是很偷税.

#include<stdio.h>
#include<cstring>
#define clear(a) memset(a, 0, sizeof(a))
#define Mercer register int
const int maxn = 100005;
double dp[maxn];
int vis[maxn], x, y, n, m;
inline const int read(){
	register int x = 0;
	register char ch = getchar();
	while(ch < '0' || ch > '9') ch = getchar();
	while(ch >= '0' && ch <= '9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();
	return x;
}
int main(){
	while(n = read() ,m = read(), n || m){
		clear(dp), clear(vis);
		for(Mercer i = 1; i <= m; ++i) x = read(), y = read(), vis[x] = y;
		for(int i = n - 1; ~i; --i)
			if(!vis[i])
				for(int j = 1; j <= 6; ++j)
					dp[i] += (dp[i+j] + 1) / 6.0;
			else dp[i] = dp[vis[i]];
		printf("%.4f\n", dp[0]);
	}
}