【HDU-oj】-1466-计算直线的交点数(DP)

本文介绍了一种通过动态规划算法来解决平面上n条直线所能产生的不同交点数目的问题。利用递推公式,可以计算出任意数量直线的所有可能交点数目组合。

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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): 9792    Accepted Submission(s): 4463


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

题解:n条直线相交的点的个数最多为C(n,2),n*(n-1)/2


我们设定一个数组a[i][j],i表示i条直线,j表示有j个交点,存在 的话a[i][j]=1


n条直线,我们假设i条平行,剩下n-i位置不定那么交点个数就是  (n-1)*i + (n-i)条直线存在的交点数,也就转换成了一个动态规划的问题


n条直线一定存在全部平行也就是交点为0的情况,即a[i][0]=1

#include <cstdio>
#include <stack>
#include <queue>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
int a[22][220];
void init()
{
	for(int i=0;i<=20;i++)
		a[i][0]=1;
	for(int n=2;n<=20;n++)
	{
		for(int i=1;i<n;i++)
		{
			for(int j=0;j<200;j++)
			{
				if(a[i][j])		// i条线有j个交点,那么令剩下(n-i)平行,则 a[n][j+i*(n-i)]成立  ,一个反过来推的过程 
					a[n][j+i*(n-i)]=1;
			}
		}		
	}	
}
int main()
{
	int n;
	init();
	while(~scanf("%d",&n))
	{
		for(int i=0;i<n*(n-1)/2;i++)
		{
			if(a[n][i])			//存在,输出交点数 
				printf("%d ",i);
		}
		printf("%d\n",n*(n-1)/2);
	}
	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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