10574 - Counting Rectangles

题目链接

题意：给定一些点，求能够成几个矩形

思路：先把点按x排序，再按y排序，然后用O(n^2)的方法找出每条垂直x轴的边，保存这些边两点的y坐标y1, y2。之后把这些边按y1排序，再按y2排序，用O(n)的方法找出有几个连续的y1, y2都相等，那么这些边两两是能构成矩形的，为C2cnt种，然后累加起来就是答案

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int N = 5005;
int t, n;
struct Point {
    int x, y;
    void read() {
    scanf("%d%d", &x, &y);
    }
    bool operator < (const Point a) const {
    if (x != a.x)
        return x < a.x;
    return y < a.y;
    }
} p[N];

struct Edge {
    int y1, y2;
    Edge(int y1 = 0, int y2 = 0) {
    this->y1 = y1;
    this->y2 = y2;
    }
    bool operator < (const Edge a) const {
    if (y1 != a.y1)
        return y1 < a.y1;
    return y2 < a.y2;
    }
} e[N * N / 2];

int main() {
    int cas = 0;
    scanf("%d", &t);
    while (t--) {
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        p[i].read();
    sort(p, p + n);
    int en = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
        if (p[j].x != p[i].x) break;
        e[en++] = Edge(p[i].y, p[j].y);
        }
    }
    sort(e, e + en);
    long long cnt = 1, ans = 0;
    for (int i = 1; i < en; i++) {
        if (e[i].y1 == e[i - 1].y1 && e[i].y2 == e[i - 1].y2)
        cnt++;
        else {
        ans += cnt * (cnt - 1) / 2;
        cnt = 1;
        }
    }
    ans += cnt * (cnt - 1) / 2;
    printf("Case %d: %lld\n", ++cas, ans);
    }
    return 0;
}

UVA 10574 - Counting Rectangles(枚举+计数)