Yangff

Microsoft Academic Graph (MAG) is a large heterogeneous graph containing entities such as authors, papers, journals, conferences and relations between them. Microsoft provides Academic Knowledge API for this contest. The Entity attributes are defined here.

m$搞了一套学术数据库的API的理论呀.

Participants are supposed to provide a REST service endpoint that can find all the 1-hop, 2-hop, and 3-hop graph paths connecting a given pair of entity identifiers in MAG. The given pair of entity identifiers could be [Id, Id], [Id, AA.AuId], [AA.AuId, Id], [AA.AuId, AA.AuId]. Each node of a path should be one of the following identifiers: Id, F.Fid, J.JId, C.CId, AA.AuId, AA.AfId. Possible edges (a pair of adjacent nodes) of a path are:

你们要给他搞一个REST服务呀，对给定的点对呀，寻找所有的1-hop, 2-hop, 3-hop路径呀。给定的点对关系可能是这样的呀[Id, Id], [Id, AA.AuId], [AA.AuId, Id], [AA.AuId, AA.AuId], 而你的路径上可以涵盖这些点Id, F.Fid, J.JId, C.CId, AA.AuId, AA.AfId

而两个点之间连边的要求是呀

Keybase proof

I hereby claim:

I am yangff on github.
I am yangff (https://keybase.io/yangff) on keybase.
I have a public key whose fingerprint is DC9F C32F 1859 DEBF B1F7 275C B424 7F16 00FE 969D

To claim this, I am signing this object:

	class Symbol
	def [](*args)
	a = [self, args, '[]']
	a.instance_eval('@_type_flag = \'[]\'')
	a
	end
	def <<(*ary)
	a = [self, ary.flatten, '<']
	a.instance_eval('@_type_flag = \'<\'')
	a

	// g++ aa.cpp -o aa.exe -std=c++14

	#include <cstdio>
	#include <functional>
	#include <tuple>

	using namespace std;

	template<int n, class Iso, class TRet, class... TArgs>
	struct _bindArgs;

	#include <cstdio>
	#include <windows.h>

	int main(){
	for (;;){
	mouse_event(MOUSEEVENTF_LEFTDOWN, 0, 0, 0, 0);
	Sleep(40);
	mouse_event(MOUSEEVENTF_LEFTUP, 0, 0, 0, 0);
	Sleep(30);
	}

	<!DOCTYPE html>
	<html lang="en">
	<head>
	<meta charset="UTF-8">
	<meta name="viewport" content="width=device-width,initial-scale=1,user-scalable=0">

	<style>
	html, body {
	height: 100%;
	-webkit-tap-highlight-color: transparent;

	void del_substr(char str, const char substr) {
	char c1, c2;
	while (*str++!=0) {
	for (c1 = str, c2 = (char)substr; c1 == c2 && c1 && *c2; c1++, c2++);
	if (*c2 == 0)
	for (c2 = str; c2++ = c1++; );
	}
	}

	void del_substr(char str, const char substr) {
	for (char c1, c2; str; str+=!!(c2)) {
	for (c1 = str, c2 = (char)substr; c1 == c2 && c1 && *c2; c1++, c2++);
	if (*c2 == 0) {
	for (c2 = str; c2 = c1; c2++, c1++);
	}
	}
	}

	void del_substr(char str, const char substr) {
	for (int i = 0, j = 0; str; str += !!(substr+j)) {
	for (j = 0; (str+i+j) == (substr+j) && *(substr+j); j++);
	str = (str+i);
	i += !(substr+j) j;
	}
	}

	void del_substr(char str, const char substr) {
	for (int i = 0, j = 0; str = (str+i); str += !!(substr+j), i += !(substr+j) * j)
	for (j = 0; (str+i+j) == (substr+j) && *(substr+j); j++);
	}