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[Linux-Unix] 3pcf DL : 0: 3PCF计算多线程实现定义：点集D、R。定义D中的点为ai∈D，R中的点为bi∈R。距离：r1、r2、r3、err 求：满足以下条件的三元组（空间中三角形）的数目 <ai, bm, bn>，|ai-bm|=r1±err且|ai-bn|=r2±err且|bm-bn|=r3±err 原始解法：对于D中每一点ai，在R中找到与之距离为r1的点集R’，找到与之距离为r2的点集R’’。在点集R’与R’’中，查找两点间距离为r3的点组数目。累加。 -3PCF Multithread computing definition: point sets D, R. The point of the definition of D ai ∈ D, R in the point of bi ∈ R. Distance: r1, r2, r3, err requirements: the triple meet the following conditions (space triangle) the number of <ai, bm, bn> , | Ai-bm | = r1 ± err and | ai-bn | = r2 ± err and | bm-bn | = r3 ± err original solution: For D, each point ai, found in R with the distance r1 point set R ' , find the point with a distance r2 set R' ' . In the point set R ' and R' ' , find the distance between two points in the point group for the number of r3. Accumulation.
Date : 2025-12-31 Size : 1kb User : Junki Lee