HDU-1466-计算直线的交点数【DP】

本文介绍了一种计算平面上n条直线(无三线共点)不同交点数的方法,并提供了一个具体的算法实现。通过递推的方式,考虑每条新增直线与已有直线的关系,包括平行与相交情况,最终输出所有可能的交点数。

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题目链接:点击打开链接

计算直线的交点数

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9518    Accepted Submission(s): 4325


Problem Description
平面上有n条直线,且无三线共点,问这些直线能有多少种不同交点数。
比如,如果n=2,则可能的交点数量为0(平行)或者1(不平行)。
 

Input
输入数据包含多个测试实例,每个测试实例占一行,每行包含一个正整数n(n<=20),n表示直线的数量.
 

Output
每个测试实例对应一行输出,从小到大列出所有相交方案,其中每个数为可能的交点数,每行的整数之间用一个空格隔开。
 

Sample Input
  
2 3
 

Sample Output
  
0 1 0 2 3

分析加入第N条直线的情况(这里以N=4为例): 
(分类方法:和第N条直线平行的在a组,其余在b组) 
1、 第四条与其余直线全部平行 :

0+4*0+0=0; 

2、 第四条与其中两条平行:

交点数为0+(n-1)*1+0=3; 

3、 第四条与其中一条平行:

这两条平行直线和另外两点直线的交点数为(n-2)*2=4,

而另外两条直线既可能平行也可能相交,

因此可能交点数为: 0+(n-2)* 2+0=4    或者  0+(n-2)*2+1=5     

4、 第四条直线不与任何一条直线平行,交点数为:0+(n-3)*3+0=3  或0+ (n-3)*3+2=5  或0+ (n-3)*3+3=6 

即n=4时,有0个,3个,4个,5个,6个不同交点数。

所以从上述n=4的分析过程中,我们发现: 

m条直线的交点方案数 =(m - r)条平行线与r条直线交叉的交点数 + r条直线本身的交点方案 = (m - r )* r + r 条之间本身的交点方案数(0 < = r < m )

将 n 条直线排成一个序列,直线 2 和直线 1 最多只有一个交点,直线 3 和直线 1 和直线 2 最多有两个交点......直线n和其他 n - 1 条直线最多有 n - 1 个交点,由此得出 n 条直线互不平行且无三线共点的最多交点数:max = 1 + 2 + . . . + (n - 1) = n * (n - 1) / 2

其中每个自由直线与每个平行直线都有一个交点,j自由直线与i平行直线的交点数为 j * i,所以n条直线的交点数等于自由直线与平行直线的交点加上自由直线内部形成的交点。

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n;
bool dp[22][222]; // dp[i][j] = true 表示 i条直线时有 j个交点为真 
void init()
{
	memset(dp,false,sizeof(dp));
	for(int i=1;i<=20;i++) // i表示直线总数 
	{
		dp[i][0]=true; // 总是存在交点为 0 的
		for(int j=1;j<i;j++) // j表示平行直线数 
		{
			int m=i-j;	// m表示自由直线数 
			for(int k=0;k<=m*(m-1)/2;k++) // m条自由直线(自由直线互不平行且无三线共点)最多形成 m*(m-1)/2个点 
			{
				if(dp[m][k])
				{
					dp[i][m*j+k]=true;
				}
			}
		}
	}
}
int main()
{
	init();
	while(~scanf("%d",&n))
	{
		printf("0");
		for(int i=1;i<=n*(n-1)/2;i++)
		{
			if(dp[n][i])
				printf(" %d",i);
		}
		puts("");
	}
	return 0;
}

### HDU 1466 Problem Description and Solution The problem **HDU 1466** involves calculating the expected number of steps to reach a certain state under specific conditions. The key elements include: #### Problem Statement Given an interactive scenario where operations are performed on numbers modulo \(998244353\), one must determine the expected number of steps required to achieve a particular outcome. For this type of problem, dynamic programming (DP) is often employed as it allows breaking down complex problems into simpler subproblems that can be solved iteratively or recursively with memoization techniques[^1]. In more detail, consider the constraints provided by similar problems such as those found in references like HDU 6327 which deals with random sequences using DP within given bounds \((1 \leq T \leq 10, 4 \leq n \leq 100)\)[^2]. These types of constraints suggest iterative approaches over small ranges might work efficiently here too. Additionally, when dealing with large inputs up to \(2 \times 10^7\) as seen in reference materials related to counting algorithms [^4], efficient data structures and optimization strategies become crucial for performance reasons. However, directly applying these methods requires understanding how they fit specifically into solving the expectation value calculation involved in HDU 1466. For instance, if each step has multiple outcomes weighted differently based on probabilities, then summing products of probability times cost across all possible states until convergence gives us our answer. To implement this approach effectively: ```python MOD = 998244353 def solve_expectation(n): dp = [0] * (n + 1) # Base case initialization depending upon problem specifics for i in range(1, n + 1): total_prob = 0 # Calculate transition probabilities from previous states for j in transitions_from(i): # Placeholder function representing valid moves prob = calculate_probability(j) next_state_cost = get_next_state_cost(j) dp[i] += prob * (next_state_cost + dp[j]) % MOD total_prob += prob dp[i] %= MOD # Normalize current state's expectation due to accumulated probability mass if total_prob != 0: dp[i] *= pow(total_prob, MOD - 2, MOD) dp[i] %= MOD return dp[n] # Example usage would depend heavily on exact rules governing transitions between states. ``` This code snippet outlines a generic framework tailored towards computing expectations via dynamic programming while adhering strictly to modular arithmetic requirements specified by the contest question format.
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