《DirectX 12 3D游戏开发实战》学习记录

《DirectX 12 3D游戏开发实战》第二章练习答案

/**
* 练习的最后两题代码比较乱，按部就班的思路，没有技巧性，
* 但也遇到不少问题，花了很长时间。
**/

#include <Windows.h>
#include <DirectXMath.h>
#include <DirectXPackedVector.h>
#include <iostream>
#include <vector>
using namespace std;
using namespace DirectX;
using namespace DirectX::PackedVector;

ostream& XM_CALLCONV operator<<(ostream& os, FXMVECTOR v)
{
XMFLOAT4 dest;
XMStoreFloat4(&dest, v);

cout << "(" << dest.x << ", " << dest.y << ", "
<< dest.z << ", " << dest.w << ")";
return os;
}

ostream& XM_CALLCONV operator<<(ostream& os, FXMMATRIX m)
{
for (int i = 0; i < 4; ++i) {
os << XMVectorGetX(m.r[i]) << "\t";
os << XMVectorGetY(m.r[i]) << "\t";
os << XMVectorGetZ(m.r[i]) << "\t";
os << XMVectorGetW(m.r[i]) << endl;
}
return os;
}

void p1()
{
XMMATRIX A(-2.0f, 0.0f, 0.0f, 0.0f,
1.0f, 3.0f, 0.0f, 0.0f,
0.0f, 0.0f, 0.0f, 0.0f,
0.0f, 0.0f, 0.0f, 0.0f);
XMMATRIX X = 1 / 2.0 * (A - 2 / 3.0 * A);
cou
3ff7
t << "X = " << endl << X << endl;
}

void p2()
{
XMMATRIX aA(-2.0f, 0.0f, 3.0f, 0.0f,
4.0f, 1.0f, -1.0f, 0.0f,
0.0f, 0.0f, 0.0f, 0.0f,
0.0f, 0.0f, 0.0f, 0.0f);
XMMATRIX aB(2.0f, -1.0f, 0.0f, 0.0f,
0.0f, 6.0f, 0.0f, 0.0f,
2.0f, -3.0f, 0.0f, 0.0f,
0.0f, 0.0f, 0.0f, 0.0f);
cout << "a. A * B = " << endl << aA * aB << endl;

XMMATRIX bA(1.0f, 2.0f, 0.0f, 0.0f,
3.0f, 4.0f, 0.0f, 0.0f,
0.0f, 0.0f, 0.0f, 0.0f,
0.0f, 0.0f, 0.0f, 0.0f);
XMMATRIX bB(-2.0f, 0.0f, 0.0f, 0.0f,
1.0f, 1.0f, 0.0f, 0.0f,
0.0f, 0.0f, 0.0f, 0.0f,
0.0f, 0.0f, 0.0f, 0.0f);
cout << "b. A * B = " << endl << bA * bB << endl;

XMMATRIX cA(2.0f, 0.0f, 2.0f, 0.0f,
0.0f, -1.0f, -3.0f, 0.0f,
0.0f, 0.0f, 1.0f, 0.0f,
0.0f, 0.0f, 0.0f, 0.0f);
XMMATRIX cB(1.0f, 0.0f, 0.0f, 0.0f,
2.0f, 0.0f, 0.0f, 0.0f,
1.0f, 0.0f, 0.0f, 0.0f,
0.0f, 0.0f, 0.0f, 0.0f);
cout << "c. A * B = " << endl << cA * cB << endl;
}

void p8()
{
XMMATRIX A(2.0f, 0.0f, 1.0f, 0.0f,
0.0f, -1.0f, -3.0f, 0.0f,
0.0f, 0.0f, 1.0f, 0.0f,
0.0f, 0.0f, 0.0f, 0.0f);
XMMATRIX B(0.5f, 0.0f, -0.5f, 0.0f,
0.0f, -1.0f, -3.0f, 0.0f,
0.0f, 0.0f, 1.0f, 0.0f,
0.0f, 0.0f, 0.0f, .0f);
XMMATRIX C = A * B;
cout << "C = A * B = " << endl << C << endl;
//C是3阶矩阵，故B是A的逆矩阵
}
//18.求转置矩阵
void transpose(vector<vector<double>> source, vector<vector<double>> &dest)
{
dest = source;
for (int i = 0; i < dest.size(); ++i) {
for (int j = i + 1; j < dest.size(); ++j) {
double temp = dest[i][j];
dest[i][j] = dest[j][i];
dest[j][i] = temp;
}
}
}
//19.求矩阵的行列式
double determinant(vector<vector<double>> a)
{
//如果不是n阶矩阵
if (a.size() != a[0].size()) {
cout << "error" << endl;
return -1;
}
//1阶矩阵直接返回首元素
if (a[0].size() == 1) return a[0][0];
else {
int n = a[0].size();

double det = 0;
//第一行第i+1列乘以其余子式
for (int i = 0; i < n; ++i) {
//创建n-1阶矩阵,初始化余子式
vector<vector<double>> temp;
for (int i = 0; i < n - 1; ++i)
temp.push_back(vector<double>());
int row = 0;
//不在第一行且不在第c+1列的元素放入temp中
for (int r = 1; r < n; ++r) {
for (int c = 0; c < n; ++c) {
if (c != i) {
temp[row].push_back(a[r][c]);
//如果当前行元素已满
if (temp[row].size() == n - 1)
++row;
}
}
}
//代数余子式，i是偶数为加，奇数为减
if (i % 2 == 0)
det += a[0][i] * determinant(temp);
else det -= a[0][i] * determinant(temp);
}
return det;
}
}
//19.求逆矩阵
void inverse(vector<vector<double>> a, vector<vector<double>> &dest)
{
double det = determinant(a);
vector<vector<double>> b;
for (int i = 0; i < a.size(); ++i)
b.push_back(vector<double>());
//求出每一行每一列的余子式
for (int i = 0; i < a.size(); ++i) {
for (int j = 0; j < a[0].size(); ++j) {
vector<vector<double>> temp;
for (int n = 0; n < a.size() - 1; ++n)
temp.push_back(vector<double>());
int row = 0;
for (int r = 0; r < a.size(); ++r) {
for (int c = 0; c < a[0].size(); ++c) {
//去除当前行和列
if (r != i && c != j) {
temp[row].push_back(a[r][c]);
if (temp[row].size() == a.size() - 1)
++row;
}
}
}
//将代数余子式填入
double m = determinant(temp);
if ((i + j) % 2 == 1 && m != 0)
m *= -1;
b[i].push_back(m / det);
}
}
//转置
transpose(b, dest);
}

int main()
{
cout.setf(ios::boolalpha);
//p1();
//p2();
//p8();
vector<vector<double>> a = { {1, 0, 0, 0}, {0, 2, 0, 0}, {0, 0, 4, 0}, {1, 2, 3, 1} };
vector<vector<double>> b;
transpose(a, b);
cout << "转置矩阵transposeA = " << endl;
for (int i = 0; i < b.size(); ++i) {
for (int j = 0; j < b[0].size(); ++j)
cout << b[i][j] << "\t";
cout << endl;
}
cout << endl;

double det = determinant(a);
cout << "行列式detA = " << det << endl << endl;

vector<vector<double>> c;
inverse(a, c);
cout << "逆矩阵inverseA = " << endl;
for (int i = 0; i < c.size(); ++i) {
for (int j = 0; j < c[0].size(); ++j)
cout << c[i][j] << "\t";
cout << endl;
}
cout << endl;

system("pause");
return 0;
}
``