uva 10574 - Counting Rectangles(计数)

题目链接：uva 10574 - Counting Rectangles

题目大意：给出n个点，问选出4个点作为定点，可以组成多少个平行与坐标轴的矩形。

解题思路：首先将点按照x排序(优化)，然后处理出所有平行于y轴的线段，记录这些线段的y1和y2，接着只要找出y1和y2值均相等的边，C(2cnt).

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
const int N = 5005;

struct point {
    int x, y;
    void get () {
        scanf("%d%d", &x, &y);
    }
    void set (int x, int y) {
        this->x = x;
        this->y = y;
    }
}p[N], l[N*N/2];
int n, m;

inline bool cmp(const point& a, const point& b) {
    if (a.x != b.x)
        return a.x < b.x;
    return a.y < b.y;
}

void init () {
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        p[i].get();

    sort(p, p + n, cmp);

    m = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i+1; j < n; j++) {
            if (p[i].x != p[j].x)
                break;

            l[m++].set(p[i].y, p[j].y);
        }
    }
    sort(l, l + m, cmp);
}

ll solve () {
    ll ans = 0;

    for (int i = 0; i < m;) {
        int mv = i+1;

        while (l[i].x == l[mv].x && l[i].y == l[mv].y)
            mv++;

        ll c = mv - i;

        if (c >= 2)
            ans += c * (c - 1) / 2;
        i = mv;
    }
    return ans;
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int i = 1; i <= cas; i++) {
        init();
        printf("Case %d: %lld\n", i, solve());
    }
    return 0;
}