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原题链接：1312.[COCI-2006_2007-contest6]#5 PROSTOR

银光蒙蒙

2026-03-19 16:04:16

发布于：江苏

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#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;

struct Rectangle {
    int x1, y1, z1, x2, y2, z2;
    
    Rectangle() : x1(0), y1(0), z1(0), x2(0), y2(0), z2(0) {}
    
    Rectangle(int a, int b, int c, int d, int e, int f) {
        x1 = min(a, d); x2 = max(a, d);
        y1 = min(b, e); y2 = max(b, e);
        z1 = min(c, f); z2 = max(c, f);
    }
};

// 优化的相交检测函数
inline bool intersect(const Rectangle& r1, const Rectangle& r2) {
    return (r1.x1 <= r2.x2 && r2.x1 <= r1.x2) &&
           (r1.y1 <= r2.y2 && r2.y1 <= r1.y2) &&
           (r1.z1 <= r2.z2 && r2.z1 <= r1.z2);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    cin >> n;
    
    vector<Rectangle> rectangles(n);
    for (int i = 0; i < n; ++i) {
        int x1, y1, z1, x2, y2, z2;
        cin >> x1 >> y1 >> z1 >> x2 >> y2 >> z2;
        rectangles[i] = Rectangle(x1, y1, z1, x2, y2, z2);
    }
    
    // 计算冲突的矩形对数
    long long count = 0;
    for (int i = 0; i < n; ++i) {
        for (int j = i + 1; j < n; ++j) {
            if (intersect(rectangles[i], rectangles[j])) {
                count++;
            }
        }
    }
    
    cout << count << "\n";
    return 0;
}

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