计算直线的交点数(set + 打表)

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): 9357    Accepted Submission(s): 4226

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:用st存一下i条直线的点数目的种类;然后枚举平行线的条数j;当前点的个数就是*iter+平行的*不平行的,画个图看下就好了;

想了下,写下就过了;有点dp的意思

代码:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
using namespace std;
set<int>st[25];
set<int>::iterator iter;
void db(){
    st[1].insert(0);
    for(int i = 2; i < 25; i++){
        st[i].insert(0);
        for(int j = 1; j < i; j++){
            for(iter = st[i - j].begin(); iter != st[i - j].end(); iter++){
                int dot = *iter + (i - j) * j;
                    st[i].insert(dot); 
            }
        }
    }
    
}
int main(){
    int n;
    db();
    while(~scanf("%d", &n)){
        for(iter = st[n].begin(); iter != st[n].end(); iter++){
            if(iter != st[n].begin())printf(" ");
            printf("%d", *iter);
        }
        puts("");
    }
    return 0;
}

 java:

package com.lanqiao.week1;

import java.util.Arrays;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Scanner;
import java.util.Set;

public class hdu1466 {
    private static Scanner cin = null;
    static Set<MyInteger>[] set = null;
    static MyInteger[][] arr = new MyInteger[25][];
    private static final int MAXN = 25;
    static{
        cin = new Scanner(System.in);
        set = new HashSet[MAXN];
    }
    static class MyInteger implements Comparable{
        int v;
        @Override
        public int hashCode() {
            return v;
        }
        public int getV() {
            return v;
        }
        public MyInteger(int v) {
            super();
            this.v = v;
        }
        @Override
        public String toString() {
            return v + "";
        }
        @Override
        public boolean equals(Object obj) {
            MyInteger t = (MyInteger)obj;
            return t.v == this.v;
        }
        @Override
        public int compareTo(Object o) {
            MyInteger t = (MyInteger)o;
            return this.v - t.getV();
        }
    }
    private static void getSet(){
        for(int i = 0; i < MAXN; i++){
            set[i] = new HashSet<MyInteger>(){};
        }
        set[1].add(new MyInteger(0));
        for(int i = 2; i < MAXN; i++){
            set[i].add(new MyInteger(0));
            for(int j = 1; j < i; j++){
                Iterator<MyInteger> iter = set[i - j].iterator();
                while(iter.hasNext()){
                    //System.out.println("zz");
                    set[i].add(new MyInteger(iter.next().getV() + j * (i - j)));
                }
            }
        }
        for(int i = 1; i < MAXN; i++){
            arr[i] = new MyInteger[set[i].size()];
            set[i].toArray(arr[i]);
            Arrays.sort(arr[i]);
        }
    }
    public static void main(String[] args) {
        getSet();
        while(cin.hasNext()){
            int N = cin.nextInt();
            
            for(int i = 0; i < arr[N].length; i++){
                if(i != 0){
                    System.out.print(" ");
                }
                System.out.print(arr[N][i].getV());
            }System.out.println();
        }
    }
}
原文地址:https://www.cnblogs.com/handsomecui/p/5475439.html