The object detector described below has been initially proposed by P.F. Felzenszwalb in [Felzenszwalb2010]. It is based on a Dalal-Triggs detector that uses a single filter on histogram of oriented gradients (HOG) features to represent an object category. This detector uses a sliding window approach, where a filter is applied at all positions and scales of an image. The first innovation is enriching the Dalal-Triggs model using a star-structured part-based model defined by a “root” filter (analogous to the Dalal-Triggs filter) plus a set of parts filters and associated deformation models. The score of one of star models at a particular position and scale within an image is the score of the root filter at the given location plus the SUM over parts of the maximum, over placements of that part, of the part filter score on its location minus a deformation cost easuring the deviation of the part from its ideal location relative to the root. Both root and part filter scores are defined by the dot product between a filter (a set of weights) and a subwindow of a feature pyramid computed from the input image. Another improvement is a representation of the class of models by a mixture of star models. The score of a mixture model at a particular position and scale is the maximum over components, of the score of that component model at the given location.
標簽： 計算機視覺
上傳時間： 2015-03-15
上傳用戶：sb_zhang
最近在學習Oracle，對測試人員而言必須掌握兩種語言：第一種是DML，數據操縱語言 (Data Manipulation Language) 是SQL語言中，負責對數據庫對象運行數據訪問工作的指令集，以INSERT、UPDATE、DELETE三種指令為核心，分別代表插入、更新與刪除。第二種是：DQL，數據查詢語言 (Data Query Language) 是SQL語言中，負責進行數據查詢而不會對數據本身進行修改的語句，這是最基本的SQL語句。核心指令為SELECT，以及一些輔助指令，如FROM、WHERE等，FROM：表示來源，可以搭配JOIN做鏈接查詢； WHERE：過濾條件；GROUP BY：在使用聚合函數時用到，如SUM,COUNT,MAX,AVG；HAVING：對聚合結果進行篩選，這是和WHERE的不同點；ORDER BY：排序。
標簽： oracle 基礎 資料
上傳時間： 2016-09-15
上傳用戶：天涯云海
批處理感知器算法的代碼matlab w1=[1,0.1,1.1;1,6.8,7.1;1,-3.5,-4.1;1,2.0,2.7;1,4.1,2.8;1,3.1,5.0;1,-0.8,-1.3; 1,0.9,1.2;1,5.0,6.4;1,3.9,4.0]; w2=[1,7.1,4.2;1,-1.4,-4.3;1,4.5,0.0;1,6.3,1.6;1,4.2,1.9;1,1.4,-3.2;1,2.4,-4.0; 1,2.5,-6.1;1,8.4,3.7;1,4.1,-2.2]; w3=[1,-3.0,-2.9;1,0.5,8.7;1,2.9,2.1;1,-0.1,5.2;1,-4.0,2.2;1,-1.3,3.7;1,-3.4,6.2; 1,-4.1,3.4;1,-5.1,1.6;1,1.9,5.1]; figure; plot(w3(:,2),w3(:,3),'ro'); hold on; plot(w2(:,2),w2(:,3),'b+'); W=[w2;-w3];%增廣樣本規范化 a=[0,0,0]; k=0;%記錄步數 n=1; y=zeros(size(W,2),1);%記錄錯分的樣本 while any(y<=0) k=k+1; y=a*transpose(W);%記錄錯分的樣本 a=a+SUM(W(find(y<=0),:));%更新a if k >= 250 break end end if k<250 disp(['a為：',num2str(a)]) disp(['k為：',num2str(k)]) else disp(['在250步以內沒有收斂，終止']) end %判決面：x2=-a2*x1/a3-a1/a3 xmin=min(min(w1(:,2)),min(w2(:,2))); xmax=max(max(w1(:,2)),max(w2(:,2))); x=xmin-1:xmax+1;%(xmax-xmin): y=-a(2)*x/a(3)-a(1)/a(3); plot(x,y)
上傳時間： 2016-11-07
上傳用戶：a1241314660
/*import java.util.Scanner; //主類 public class student122 { //主方法 public static void main(String[] args){ //定義7個元素的字符數組 String[] st = new String[7]; inputSt(st); //調用輸入方法 calculateSt(st); //調用計算方法 outputSt(st); //調用輸出方法 } //其他方法 //輸入方法 private static void inputSt(String st[]){ System.out.println("輸入學生的信息："); System.out.println("學號 姓名 成績1，2，3"); //創建鍵盤輸入類 Scanner ss = new Scanner(System.in); for(int i=0; i<5; i++){ st[i] = ss.next(); //鍵盤輸入1個字符串 } } //計算方法 private static void calculateSt(String[] st){ int SUM = 0; //總分賦初值 int ave = 0; //平均分賦初值 for(int i=2;i<5;i++) { /計總分，字符變換成整數后進行計算 SUM += Integer.parseInt(st[i]); } ave = SUM/3; //計算平均分 //整數變換成字符后保存到數組里 st[5] = String.valueOf(SUM); st[6] = String.valueOf(ave); } //輸出方法 private static void outputSt(String[] st){ System.out.print("學號 姓名 "); //不換行 System.out.print("成績1 成績2 成績3 "); System.out.println("總分 平均分");//換行 //輸出學生信息 for(int i=0; i<7; i++){ //按格式輸出，小于6個字符，補充空格 System.out.printf("%6s", st[i]); } System.out.println(); //輸出換行 } }*/ import java.util.Scanner; public class student122 { public static void main(String[] args) { // TODO 自動生成的方法存根 String[][] st = new String[3][8]; inputSt(st); calculateSt(st); outputSt(st); } //輸入方法 private static void inputSt(String st[][]) { System.out.println("輸入學生信息："); System.out.println("班級 學號 姓名 成績：數學 物理 化學"); //創建鍵盤輸入類 Scanner ss = new Scanner(System.in); for(int j = 0; j < 3; j++) { for(int i = 0; i < 6; i++) { st[j][i] = ss.next(); } } } //輸出方法 private static void outputSt(String st[][]) { System.out.println("序號 班級 學號 姓名 成績：數學 物理 化學 總分 平均分"); //輸出學生信息 for(int j = 0; j < 3; j++) { System.out.print(j+1 + ":"); for(int i = 0; i < 8; i++) { System.out.printf("%6s", st[j][i]); } System.out.println(); } } //計算方法 private static void calculateSt(String[][] st) { int SUM1 = 0; int SUM2 = 0; int SUM3 = 0; int ave1 = 0; int ave2 = 0; int ave3 = 0; for(int i = 3; i < 6; i++) { SUM1 += Integer.parseInt(st[0][i]); } ave1 = SUM1/3; for(int i = 3; i < 6; i++) { SUM2 += Integer.parseInt(st[1][i]); } ave2 = SUM2/3; for(int i = 3; i < 6; i++) { SUM3 += Integer.parseInt(st[2][i]); } ave3 = SUM3/3; st[0][6] = String.valueOf(SUM1); st[1][6] = String.valueOf(SUM2); st[2][6] = String.valueOf(SUM3); st[0][7] = String.valueOf(ave1); st[1][7] = String.valueOf(ave2); st[2][7] = String.valueOf(ave3); } }
上傳時間： 2017-03-17
上傳用戶：simple
1．Describe a Θ(n lg n)-time algorithm that, given a set S of n integers and another integer x, determines whether or not there exist two elements in S whose SUM is exactly x. (Implement exercise 2.3-7.)
上傳時間： 2017-04-01
上傳用戶：糖兒水嘻嘻
1．Describe a Θ(n lg n)-time algorithm that, given a set S of n integers and another integer x, determines whether or not there exist two elements in S whose SUM is exactly x. (Implement exercise 2.3-7.) #include<stdio.h> #include<stdlib.h> void merge(int arr[],int low,int mid,int high){ int i,k; int *tmp=(int*)malloc((high-low+1)*sizeof(int)); int left_low=low; int left_high=mid; int right_low=mid+1; int right_high=high; for(k=0;left_low<=left_high&&right_low<=right_high;k++) { if(arr[left_low]<=arr[right_low]){ tmp[k]=arr[left_low++]; } else{ tmp[k]=arr[right_low++]; } } if(left_low<=left_high){ for(i=left_low;i<=left_high;i++){ tmp[k++]=arr[i]; } } if(right_low<=right_high){ for(i=right_low;i<=right_high;i++) tmp[k++]=arr[i]; } for(i=0;i<high-low+1;i++) arr[low+i]=tmp[i]; } void merge_sort(int a[],int p,int r){ int q; if(p<r){ q=(p+r)/2; merge_sort(a,p,q); merge_sort(a,q+1,r); merge(a,p,q,r); } } int main(){ int a[8]={3,5,8,6,4,1,1}; int i,j; int x=10; merge_sort(a,0,6); printf("after Merging-Sort:\n"); for(i=0;i<7;i++){ printf("%d",a[i]); } printf("\n"); i=0;j=6; do{ if(a[i]+a[j]==x){ printf("exist"); break; } if(a[i]+a[j]>x) j--; if(a[i]+a[j]<x) i++; }while(i<=j); if(i>j) printf("not exist"); system("pause"); return 0; }
上傳時間： 2017-04-01
上傳用戶：糖兒水嘻嘻
function [alpha,N,U]=youxianchafen2(r1,r2,up,under,num,deta) %[alpha,N,U]=youxianchafen2(a,r1,r2,up,under,num,deta) %該函數用有限差分法求解有兩種介質的正方形區域的二維拉普拉斯方程的數值解 %函數返回迭代因子、迭代次數以及迭代完成后所求區域內網格節點處的值 %a為正方形求解區域的邊長 %r1,r2分別表示兩種介質的電導率 %up,under分別為上下邊界值 %num表示將區域每邊的網格剖分個數 %deta為迭代過程中所允許的相對誤差限 n=num+1; %每邊節點數 U(n,n)=0; %節點處數值矩陣 N=0; %迭代次數初值 alpha=2/(1+sin(pi/num));%超松弛迭代因子 k=r1/r2; %兩介質電導率之比 U(1,1:n)=up; %求解區域上邊界第一類邊界條件 U(n,1:n)=under; %求解區域下邊界第一類邊界條件 U(2:num,1)=0;U(2:num,n)=0; for i=2:num U(i,2:num)=up-(up-under)/num*(i-1);%采用線性賦值對上下邊界之間的節點賦迭代初值 end G=1; while G>0 %迭代條件：不滿足相對誤差限要求的節點數目G不為零 Un=U; %完成第n次迭代后所有節點處的值 G=0; %每完成一次迭代將不滿足相對誤差限要求的節點數目歸零 for j=1:n for i=2:num U1=U(i,j); %第n次迭代時網格節點處的值 if j==1 %第n+1次迭代左邊界第二類邊界條件 U(i,j)=1/4*(2*U(i,j+1)+U(i-1,j)+U(i+1,j)); end if (j>1)&&(j U2=1/4*(U(i,j+1)+ U(i-1,j)+ U(i,j-1)+ U(i+1,j)); U(i,j)=U1+alpha*(U2-U1); %引入超松弛迭代因子后的網格節點處的值 end if i==n+1-j %第n+1次迭代兩介質分界面（與網格對角線重合）第二類邊界條件 U(i,j)=1/4*(2/(1+k)*(U(i,j+1)+U(i+1,j))+2*k/(1+k)*(U(i-1,j)+U(i,j-1))); end if j==n %第n+1次迭代右邊界第二類邊界條件 U(i,n)=1/4*(2*U(i,j-1)+U(i-1,j)+U(i+1,j)); end end end N=N+1 %顯示迭代次數 Un1=U; %完成第n+1次迭代后所有節點處的值 err=abs((Un1-Un)./Un1);%第n+1次迭代與第n次迭代所有節點值的相對誤差 err(1,1:n)=0; %上邊界節點相對誤差置零 err(n,1:n)=0; %下邊界節點相對誤差置零 G=SUM(SUM(err>deta))%顯示每次迭代后不滿足相對誤差限要求的節點數目G end
標簽： 有限差分
上傳時間： 2018-07-13
上傳用戶：Kemin
# include<stdio.h> # include<math.h> # define N 3 main(){ float NF2(float *x,float *y); float A[N][N]={{10,-1,-2},{-1,10,-2},{-1,-1,5}}; float b[N]={7.2,8.3,4.2},SUM=0; float x[N]= {0,0,0},y[N]={0},x0[N]={}; int i,j,n=0; for(i=0;i<N;i++) { x[i]=x0[i]; } for(n=0;;n++){ //計算下一個值 for(i=0;i<N;i++){ SUM=0; for(j=0;j<N;j++){ if(j!=i){ SUM=SUM+A[i][j]*x[j]; } } y[i]=(1/A[i][i])*(b[i]-SUM); //SUM=0; } //判斷誤差大小 if(NF2(x,y)>0.01){ for(i=0;i<N;i++){ x[i]=y[i]; } } else break; } printf("經過%d次雅可比迭代解出方程組的解：\n",n+1); for(i=0;i<N;i++){ printf("%f ",y[i]); } } //求兩個向量差的二范數函數 float NF2(float *x,float *y){ int i; float z,SUM1=0; for(i=0;i<N;i++){ SUM1=SUM1+pow(y[i]-x[i],2); } z=sqrt(SUM1); return z; }
上傳時間： 2019-10-13
上傳用戶：大萌萌撒
A wireless communication network can be viewed as a collection of nodes, located in some domain, which can in turn be transmitters or receivers (depending on the network considered, nodes may be mobile users, base stations in a cellular network, access points of a WiFi mesh etc.). At a given time, several nodes transmit simultaneously, each toward its own receiver. Each transmitter–receiver pair requires its own wireless link. The signal received from the link transmitter may be jammed by the signals received from the other transmitters. Even in the simplest model where the signal power radiated from a point decays in an isotropic way with Euclidean distance, the geometry of the locations of the nodes plays a key role since it determines the signal to interference and noise ratio (SINR) at each receiver and hence the possibility of establishing simultaneously this collection of links at a given bit rate. The interference seen by a receiver is the SUM of the signal powers received from all transmitters, except its own transmitter.
標簽： Stochastic Geometry Networks Wireless Volume and II
上傳時間： 2020-06-01
上傳用戶：shancjb
【例3.1]4位全加器module adder 4（cout，SUM i na，i nb，cin）；output[3：0]SUM output cout；input[3：0]i na，i nb；input cin；assign（cout，SUMl=i na +i nb+ci n；endmodule【例3.2]4位計數器module count 4（out，reset，clk）；output[3：0]out；input reset，cl k；regl 3：01 out；always@posedge clk）
標簽： verilog
上傳時間： 2022-06-16
上傳用戶：canderile
