Codestorm：Counting Triangles 查各种三角形的个数

题目链接：https://www.hackerrank.com/contests/codestorm/challenges/ilia

这周六玩了一天的Codestorm，这个题目是真的很好玩，无奈只做出了四道题，自己太菜，difficult的题目一道题都没出，把moderate的题目拿出来总结一下吧。
给了一些棍子，每根棍子的长度各不相同，然后问这些棍子组成的锐角三角形的个数、直角三角形的个数、钝角三角形的个数。
思路很明显，枚举前面两根木棒的长度，然后二分第三根木棒的长度。复杂度O(n^2logn)。

代码：

#include <iostream>  
#include <algorithm>  
#include <cmath>  
#include <vector>  
#include <string>  
#include <cstring>  
#pragma warning(disable:4996)  
using namespace std;

#define maxn 10005

typedef long long ll;

ll n;
ll val_2[maxn];
ll val[maxn];
ll ha[maxn];

int main()
{
	//freopen("i.txt", "r", stdin);
	//freopen("o.txt", "w", stdout);

	ll i, j, pos, v, v_sum;
	ll cnt1, cnt2, cnt3, pos1, pos2, pos3, pos_sum;
	memset(ha, 0, sizeof(ha));

	scanf("%lld", &n);
	for (i = 1; i <= n; i++)
	{
		scanf("%lld", &val[i]);
		val_2[i] = val[i] * val[i];
		ha[val[i]]++;
	}
	sort(val + 1, val + n + 1);
	sort(val_2 + 1, val_2 + n + 1);
	cnt1 = 0;
	cnt2 = 0;
	cnt3 = 0;
	for (i = 1; i <= n; i++)
	{
		for (j = i + 1; j <= n; j++)
		{
			v = val_2[i] + val_2[j];
			pos = lower_bound(val_2 + j + 1, val_2 + n + 1, v) - val_2;

			if (val_2[pos] == v)
			{
				cnt2 = cnt2 + ha[val[pos]];
				pos3 = pos + ha[val[pos]];
			}
			else
			{
				pos3 = pos;
			}
			pos1 = pos - (j + 1);
	
			v_sum = val[i] + val[j];
			pos_sum = lower_bound(val + j + 1, val + n + 1, v_sum) - val;

			pos2 = pos_sum - pos3;

			if (pos1 >= 0)
				cnt1 = cnt1 + pos1;
			if (pos2 >= 0)
				cnt3 = cnt3 + pos2;
		}
	}
	cout << cnt1 << " " << cnt2 << " " << cnt3 << endl;
	//system("pause");
	return 0;
}

亮点在后面，题解的思想是，如果对每根棍子按长度排好序之后，用sliding window的思想能够把时间复杂度降到O(n^2)。看了一下这个代码，觉得写得更好，充分利用了前面比较的关系。
代码：

#include <iostream>    
#include <algorithm>    
#include <cmath>    
#include <vector>    
#include <string>    
#include <cstring>    
#pragma warning(disable:4996)    
using namespace std;  
  
#define MA(x,y) ((x)>(y)?(x):(y))  
  
const int N = 5002;  
int a[N], b[N], n;  
  
int input()   
{  
    //freopen("i.txt", "r", stdin);  
    //freopen("o.txt", "w", stdout);  
  
    scanf("%d", &n);  
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i] * a[i];  
    return 0;  
}  
  
int sol()   
{  
    long long x = 0, y = 0, z = 0;  
  
    sort(a + 1, a + n + 1);  
    sort(b + 1, b + n + 1);  
  
    for (int i = 1; i <= n; i++)   
    {  
        int p = i + 1;  
        int q = i + 1;  
        for (int j = i + 1; j <= n; j++)   
        {  
            while (p<n && b[i] + b[j] >= b[p + 1])   
                p++;  
            q = MA(q, p);  
            while (q<n && a[i] + a[j]>a[q + 1])   
                q++;  
            if (b[i] + b[j] == b[p])  
            {  
                x += MA(p - 1 - j, 0);  
                y++;  
                z += q - p;  
            }  
            else   
            {  
                x += MA(p - j, 0);  
                z += q - p;  
            }  
        }  
    }  
    cout << x << " " << y << " " << z << endl;  
    return 0;  
}  
  
int main()  
{  
    input();  
    sol();  
    return 0;  
}