91av视频/亚洲h视频/操亚洲美女/外国一级黄色毛片 - 国产三级三级三级三级

  • 大小: 569KB
    文件類型: .zip
    金幣: 2
    下載: 0 次
    發布日期: 2021-05-13
  • 語言: 其他
  • 標簽: 聚類??數據挖掘??

資源簡介

本實驗利用兩類數據:模擬數據與真實數據。模擬數據有著名復雜網絡學者Mark Newmann所提出,該網絡包括128個節點,每個節點的度為16,網絡包含4個社團結構,每個社團包含32個節點,每個節點與社團內部節點有k1個節點相互鏈接,與社團外部有k2個節點相互鏈接 (k1+k2=16)。通過調節參數k2 (k2=1,2,3,4,5,6,7,8)增加社團構建檢測難度。http://www-personal.umich.edu/~mejn/ 真實數據集:跆拳道俱樂部數據由34個節點組成,由于管理上的分歧,俱樂部分解成兩個社團。 包括代碼和文檔

資源截圖

代碼片段和文件信息

#include?
#include?
#include?
#include?
#include?
#define?random(x)?(rand()%(x))
using?namespace?std;
int?edge?=?129;
int?maze[129][129]?=?{0};
int?N[129]?=?{0};
double?s[129][129]?=?{0};
int?pt[129]?=?{0};



void?similarity(int?a?int?b?double?arr[][129]?int?sp)
{
????srand((int)time(0));//隨機種子
????int?tmp?=?0;
????int?mx?=?0;//相似度最大的坐標
????//int?sp?=?random(a-1)?+?1;?//初始點
????int?threshold?=?1?;?//閾值
????cout?<????while(threshold++?<=?32){
????????tmp?=?0;
????????mx?=?0;
??????for(int?i?=?1;?i?????????if(arr[sp][i]?>?tmp){
????????????tmp?=?arr[sp][i];
????????????mx?=?i;
????????}
??????}
??????if(mx?==?0)
????????break;
??????if(pt[mx]?==?0){
????????cout?<????????pt[mx]?=?1;
??????}
??????else{
????????threshold--;
??????}
??????arr[sp][mx]?=?-1;
??????for(int?i?=?1;?i?????????if(arr[sp][i]?>=?0?&&?arr[sp][i]???????????arr[sp][i]?=?arr[mx][i];
????????}
????????arr[mx][i]?=?0;
??????}
????}
}

int?mxedge[35]?=?{0013610152128364555667891105120136153171190210231253276300325351378406435465496528561};

void?density(int?a?int?b?int?arr[][129]?int?sp)
{
?????srand((int)time(0));//隨機種子
?????//int?sp?=?random(a-1)?+?1;?//初始點
?????int?threshold?=?1?;?//閾值
?????int?tmp?=?0;
?????int?mx?=?0;
?????int?cnt?=?0;
?????vector?gp;
?????gp.push_back(sp);
?????cout?<?????while(threshold++?<=?32)?{
????????tmp?=?0;
????????for(int?i?=?1;?i?????????????cnt?=?0;
????????????if(arr[sp][i]?!=?-1?&&?i?!=?sp)?{
????????????????for(int?j?=?0;?j?????????????????????if(arr[i][gp[j]]?==?1)
????????????????????????cnt++;
????????????????}
????????????}
????????????if(cnt?>?tmp){
????????????????tmp?=?cnt;
????????????????mx?=?i;
????????????}
????????}
????????if(mx?==?0)?{
????????????break;
????????}
????????if(pt[mx]?==?0){
????????????cout?<????????????pt[mx]?=?1;
????????}
????????else{
????????????threshold--;
????????}
????????arr[sp][mx]?=?-1;
????????gp.push_back(mx);
?????}
}

void?pre()
{
????int?a?b;
????double?ad;
????FILE?*fp?=?fopen(“data2.txt“?“r“);
????if(fp?==?NULL)?{
????????cout?<????}
????while(!feof(fp)){
????????fscanf(fp?“%d%d“?&a?&b);
????????maze[a][b]?=?1;
????????maze[b][a]?=?1;
????????N[a]++;
????????N[b]++;
????}
????fclose(fp);
????for(int?i?=?1;?i?????????for(int?j?=?i+1;?j?????????????ad?=?0;
????????????for(int?z?=?1;?z?????????????????if(maze[i][z]?==?1?&&?maze[j][z]?==?1){
????????????????????ad++;
????????????????}
????????????}
????????????s[i][j]?=?ad?/?(N[i]?+?N[j]?-?ad);
????????}
????}
}

int?main()
{
????pre();
????/*for(int?i?=?1;?i?????????for(int?j?=?1;?j?????????????printf(“%.2f?“s[i

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件???????14998??2018-07-05?13:40??16030147001-錢宇珊-第3題\data1.txt
?????文件???????15046??2018-07-05?14:16??16030147001-錢宇珊-第3題\data2.txt
?????文件????????3537??2018-07-05?14:28??16030147001-錢宇珊-第3題\julei.cpp
?????文件?????????480??2018-06-23?21:58??16030147001-錢宇珊-第3題\karate.txt
?????文件????????2509??2018-07-05?14:16??16030147001-錢宇珊-第3題\makedata.cpp
?????文件??????591496??2018-07-05?15:39??16030147001-錢宇珊-第3題\數據挖掘第二次.docx
?????目錄???????????0??2018-07-05?15:41??16030147001-錢宇珊-第3題\

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