【问题描述】 在一个N*N的点阵中，如N=4,你现在站在（1，1），出口在（4，4）。你可以通过上、下、左、右四种移动方法，在迷宫内行走，但是同一个位置不可以访问两次，亦不可以越界。表格最上面的一行加黑数字A[1..4]分别表示迷宫第I列中需要访问并仅可以访问的格子数。右边一行加下划线数字B[1..4]则表示迷宫第I行需要访问并仅可以访问的格子数。如图中带括号红色数字就是一条符合条件的路线。 给定N，A[1..N] B[1..N]。输出一条符合条件的路线，若无解，输出NO ANSWER。（使用U，D，L，R分别表示上、下、左、右。） 2 2 1 2 （4，4） 1 （2，3） （3，3） （4，3） 3 （1，2） （2，2） 2 （1，1） 1 【输入格式】 第一行是数m (n < 6 )。第二行有n个数，表示a[1]..a[n]。第三行有n个数，表示b[1]..b[n]。 【输出格式】 仅有一行。若有解则输出一条可行路线，否则输出“NO ANSWER”。
标签： 点阵
上传时间： 2014-06-21
上传用户：llandlu
实验源代码 //Warshall.cpp #include<stdio.h> void warshall(int k,int n) { int i , j, t; int temp[20][20]; for(int a=0;a<k;a++) { printf("请输入矩阵第%d 行元素:",a); for(int b=0;b<n;b++) { scanf ("%d",&temp[a][b]); } } for(i=0;i<k;i++){ for( j=0;j<k;j++){ if(temp[ j][i]==1) { for(t=0;t<n;t++) { temp[ j][t]=temp[i][t]||temp[ j][t]; } } } } printf("可传递闭包关系矩阵是:\n"); for(i=0;i<k;i++) { for( j=0;j<n;j++) { printf("%d", temp[i][ j]); } printf("\n"); } } void main() { printf("利用 Warshall 算法求二元关系的可传递闭包\n"); void warshall(int,int); int k , n; printf("请输入矩阵的行数 i: "); scanf("%d",&k); 四川大学实验报告 printf("请输入矩阵的列数 j: "); scanf("%d",&n); warshall(k,n); }
上传时间： 2016-06-27
上传用户：梁雪文以
#include "iostream" using namespace std; class Matrix { private: double** A; //矩阵A double *b; //向量b public: int size; Matrix(int ); ~Matrix(); friend double* Dooli(Matrix& ); void Input(); void Disp(); }; Matrix::Matrix(int x) { size=x; //为向量b分配空间并初始化为0 b=new double [x]; for(int j=0;j<x;j++) b[j]=0; //为向量A分配空间并初始化为0 A=new double* [x]; for(int i=0;i<x;i++) A[i]=new double [x]; for(int m=0;m<x;m++) for(int n=0;n<x;n++) A[m][n]=0; } Matrix::~Matrix() { cout<<"正在析构中~~~~"<<endl; delete b; for(int i=0;i<size;i++) delete A[i]; delete A; } void Matrix::Disp() { for(int i=0;i<size;i++) { for(int j=0;j<size;j++) cout<<A[i][j]<<" "; cout<<endl; } } void Matrix::Input() { cout<<"请输入A:"<<endl; for(int i=0;i<size;i++) for(int j=0;j<size;j++){ cout<<"第"<<i+1<<"行"<<"第"<<j+1<<"列:"<<endl; cin>>A[i][j]; } cout<<"请输入b:"<<endl; for(int j=0;j<size;j++){ cout<<"第"<<j+1<<"个:"<<endl; cin>>b[j]; } } double* Dooli(Matrix& A) { double *Xn=new double [A.size]; Matrix L(A.size),U(A.size); //分别求得U，L的第一行与第一列 for(int i=0;i<A.size;i++) U.A[0][i]=A.A[0][i]; for(int j=1;j<A.size;j++) L.A[j][0]=A.A[j][0]/U.A[0][0]; //分别求得U，L的第r行，第r列 double temp1=0,temp2=0; for(int r=1;r<A.size;r++){ //U for(int i=r;i<A.size;i++){ for(int k=0;k<r-1;k++) temp1=temp1+L.A[r][k]*U.A[k][i]; U.A[r][i]=A.A[r][i]-temp1; } //L for(int i=r+1;i<A.size;i++){ for(int k=0;k<r-1;k++) temp2=temp2+L.A[i][k]*U.A[k][r]; L.A[i][r]=(A.A[i][r]-temp2)/U.A[r][r]; } } cout<<"计算U得:"<<endl; U.Disp(); cout<<"计算L的:"<<endl; L.Disp(); double *Y=new double [A.size]; Y[0]=A.b[0]; for(int i=1;i<A.size;i++ ){ double temp3=0; for(int k=0;k<i-1;k++) temp3=temp3+L.A[i][k]*Y[k]; Y[i]=A.b[i]-temp3; } Xn[A.size-1]=Y[A.size-1]/U.A[A.size-1][A.size-1]; for(int i=A.size-1;i>=0;i--){ double temp4=0; for(int k=i+1;k<A.size;k++) temp4=temp4+U.A[i][k]*Xn[k]; Xn[i]=(Y[i]-temp4)/U.A[i][i]; } return Xn; } int main() { Matrix B(4); B.Input(); double *X; X=Dooli(B); cout<<"~~~~解得:"<<endl; for(int i=0;i<B.size;i++) cout<<"X["<<i<<"]:"<<X[i]<<" "; cout<<endl<<"呵呵呵呵呵"; return 0; }
标签： 道理特分解法
上传时间： 2018-05-20
上传用户：Aa123456789
n present power system, the engineers face variety of challenges in planning, construction and operation. In some of the problems, the engineers need to use managerial talents. In system design or upgrading the entire system into automatic control instead of slow response of human operator, the engineers need to exercise more technical knowledge and experience. It is principally the engi- neer’s ability to achieve the success in all respect and provide the reliable and uninterrupted service to the customers. This chapter covers some important areas of the traditional power system that helps engineers to overcome the challenges. It emphasizes the characteristics of the various components of a power system such as generation, transmission, distribution, protection and SCADA system. During normal operating conditions and disturbances, the acquired knowledge will pro- vide the engineers the ability to analyse the performance of the complex system and execute future improvement
上传时间： 2020-06-07
上传用户：shancjb
The basic topic of this book is solving problems from system and control theory using convex optimization. We show that a wide variety of problems arising in system and control theory can be reduced to a handful of standard convex and quasiconvex optimization problems that involve matrix inequalities. For a few special cases there are “analytic solutions” to these problems, but our main point is that they can be solved numerically in all cases. These standard problems can be solved in polynomial- time (by, e.g., the ellipsoid algorithm of Shor, Nemirovskii, and Yudin), and so are tractable, at least in a theoretical sense. Recently developed interior-point methods for these standard problems have been found to be extremely efficient in practice. Therefore, we consider the original problems from system and control theory as solved.
标签： Linear_Matrix_Inequalities_in_Sys tem
上传时间： 2020-06-10
上传用户：shancjb
概览CM ISP(Cortex-M In-System Programmer)是为华大半导体(HDSC)的Cortex-M 系列MCU 提供的一款在线编程器软件，支持华大半导体旗下所有的Cortex-M 系列MCU 产品。本文将介绍在线编程器软件（HDSC.exe）的使用方法和编程注意事项。本文适用于在线编程器软件版本号为V2.02。
上传时间： 2022-06-25
上传用户：kingwide
AR0231AT7C00XUEA0-DRBR（RGB滤光）安森美半导体推出采用突破性减少LED闪烁 (LFM)技术的新的230万像素CMOS图像传感器样品AR0231AT，为汽车先进驾驶辅助系统(ADAS)应用确立了一个新基准。新器件能捕获1080p高动态范围(HDR)视频，还具备支持汽车安全完整性等级B(ASIL B)的特性。LFM技术(专利申请中)消除交通信号灯和汽车LED照明的高频LED闪烁，令交通信号阅读算法能于所有光照条件下工作。AR0231AT具有1/2.7英寸(6.82 mm)光学格式和1928(水平) x 1208(垂直)有源像素阵列。它采用最新的3.0微米背照式(BSI)像素及安森美半导体的DR-Pix™技术，提供双转换增益以在所有光照条件下提升性能。它以线性、HDR或LFM模式捕获图像，并提供模式间的帧到帧情境切换。 AR0231AT提供达4重曝光的HDR，以出色的噪声性能捕获超过120dB的动态范围。AR0231AT能同步支持多个摄相机，以易于在汽车应用中实现多个传感器节点，和通过一个简单的双线串行接口实现用户可编程性。它还有多个数据接口，包括MIPI(移动产业处理器接口)、并行和HiSPi(高速串行像素接口)。其它关键特性还包括可选自动化或用户控制的黑电平控制，支持扩频时钟输入和提供多色滤波阵列选择。封装和现状：AR0231AT采用11 mm x 10 mm iBGA-121封装，现提供工程样品。工作温度范围为-40℃至105℃(环境温度)，将完全通过AEC-Q100认证。
标签： 图像传感器
上传时间： 2022-06-27
上传用户：XuVshu
This example shows how you can use signal functions in the Visiondebugger to simulate a signal that is coming into one of the analog inputs of the LPC21xx.The Measure example is described in detail in the Getting StartedUser's Guide.The MEASURE example program is available for several targets:Simulator: uVision Simulator for LPC2129MCB2100: Keil MCB2100 evaluation board with ULINK debugger - Application is loaded to internal Flash. - Switch S2 (INT1) is used as GPIO and sampled (jumper positions: J1= off, J7= on) - potentiometer POT1 is sampled as AIN0 (jumper position: J2= on) - serial port COM1 parameters: 9600 baud, no parity, 8-bits, 1 stop bit, flow control noneMCB2130: Keil MCB2130 evaluation board with ULINK debugger - Application is loaded to internal Flash. - Switch S2 (INT1) is used as GPIO and sampled (jumper positions: J1= off, J7= on) - potentiometer POT1 is sampled as AIN1 (jumper position: J2= on) - serial port COM1 parameters: 9600 baud, no parity, 8-bits, 1 stop bit, flow control none
标签： dac8568
上传时间： 2022-06-28
上传用户：
VME has been the de-facto bus standard for Commercial off the Shelf(COTS ) Circuit Card Assemblies since the 1980's. VME boards have proven to be remarkably capable of evolving to support newer technologies with innovations such as VME Subsystem Bus, PCI Mezzanine Cards(PMC's) and VME320.However, advances in technologies, appearing particularly in interconnects, have demonstratedthe need for an advance in system development. This advance needs to accommodate high speed interconnect, particularly serial interconnects, and higher power delivery in concert with better heat removal.
标签： vpx标准
上传时间： 2022-07-27
上传用户：bluedrops
Specification for conveying ITU-R System B Teletext in DVB bitstreams
标签： Specification bitstreams conveying Teletext
上传时间： 2014-02-18
上传用户：mikesering
