矩阵链乘法打印矩阵序列
mtrix chain multiplication print the sequence of the mattrices
我用c++编写了动态规划中矩阵链乘法的代码。
打印正确的矩阵括号的递归调用中存在错误。我正在从文本文件中获取输入并在文本文件上提供输出。请帮助..
#include <iostream>
#include <fstream>
#include <limits.h>
using namespace std;
int * MatrixChainOrder(int p[], int n)
{
static int m[100][100];
static int s[100][100];
int j, q;
int min = INT_MAX;
for (int i = 1; i <= n; i++)
m[i][i] = 0;
for (int L = 2; L <= n; L++) {
for (int i = 1; i <= n - L + 1; i++) {
j = i + L - 1;
m[i][j] = min;
for (int k = i; k <= j - 1; k++) {
q = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j];
if (q < m[i][j]) {
m[i][j] = q;
s[i][j] = k;
}
}
}
}
return (*s);
}
void Print(int *s, int i, int j)
{
ofstream outfile("output.text");
if (i == j)
{
outfile << "a1";
}
else
outfile << "(";
{
Print(*s, i, s[i][j]);
Print(*s, s[i][j] + 1, j);
outfile << ")";
}
outfile.close();
}
int main()
{
int arr[100];
int num, i = 0;
ifstream infile("input.text");
while (infile)
{
infile >> num;
arr[i] = num;
i++;
}
i = i - 1;
infile.close();
Print(MatrixChainOrder(arr, i - 1), 0, i - 1);
return 0;
}
在 C++ 中,最好对数组使用 std::vector
。除此之外,您不能像那样混合使用指针和数组,因为编译器会丢失数组大小。
例如这不起作用:
int x[10][20];
void foo(int *ptr)
{
//the numbers 10 and 20 have not been passed through
}
不过你可以改成
int x[10][20];
void foo(int arr[10][20])
{
//the numbers 10 and 20 are available
}
MatrixChainOrder
应该是 return 一个数字,根据这个 link
int MatrixChainOrder(int s[100][100], int p[], int n)
{
int m[100][100];
for (int i = 0; i < 100; i++) m[i][i] = 0;
for (int i = 0; i < 100; i++) s[i][i] = 0;
int q = 0;
for (int L = 2; L <= n; L++) {
for (int i = 1; i <= n - L + 1; i++) {
int j = i + L - 1;
m[i][j] = INT_MAX;
for (int k = i; k <= j - 1; k++) {
q = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j];
if (q < m[i][j]) {
m[i][j] = q;
s[i][j] = k;
}
}
}
}
return q;
}
int main()
{
int arr[] = { 40, 20, 30, 10, 30 };
int array_size = sizeof(arr) / sizeof(int);
int n = array_size - 1;
int s[100][100];
int minimum = MatrixChainOrder(s, arr, n);
printf("{ 40, 20, 30, 10, 30 } should result in 26000 : %d\n", minimum);
return 0;
}
同样你可以改变你的Print
函数
void Print(int s[100][100], int i, int j)
{
if (i < 0 || i >= 100 || j < 0 || j >= 100)
{
cout << "array bound error\n";
}
//safely access s[i][j] ...
}
我用c++编写了动态规划中矩阵链乘法的代码。 打印正确的矩阵括号的递归调用中存在错误。我正在从文本文件中获取输入并在文本文件上提供输出。请帮助..
#include <iostream>
#include <fstream>
#include <limits.h>
using namespace std;
int * MatrixChainOrder(int p[], int n)
{
static int m[100][100];
static int s[100][100];
int j, q;
int min = INT_MAX;
for (int i = 1; i <= n; i++)
m[i][i] = 0;
for (int L = 2; L <= n; L++) {
for (int i = 1; i <= n - L + 1; i++) {
j = i + L - 1;
m[i][j] = min;
for (int k = i; k <= j - 1; k++) {
q = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j];
if (q < m[i][j]) {
m[i][j] = q;
s[i][j] = k;
}
}
}
}
return (*s);
}
void Print(int *s, int i, int j)
{
ofstream outfile("output.text");
if (i == j)
{
outfile << "a1";
}
else
outfile << "(";
{
Print(*s, i, s[i][j]);
Print(*s, s[i][j] + 1, j);
outfile << ")";
}
outfile.close();
}
int main()
{
int arr[100];
int num, i = 0;
ifstream infile("input.text");
while (infile)
{
infile >> num;
arr[i] = num;
i++;
}
i = i - 1;
infile.close();
Print(MatrixChainOrder(arr, i - 1), 0, i - 1);
return 0;
}
在 C++ 中,最好对数组使用 std::vector
。除此之外,您不能像那样混合使用指针和数组,因为编译器会丢失数组大小。
例如这不起作用:
int x[10][20];
void foo(int *ptr)
{
//the numbers 10 and 20 have not been passed through
}
不过你可以改成
int x[10][20];
void foo(int arr[10][20])
{
//the numbers 10 and 20 are available
}
MatrixChainOrder
应该是 return 一个数字,根据这个 link
int MatrixChainOrder(int s[100][100], int p[], int n)
{
int m[100][100];
for (int i = 0; i < 100; i++) m[i][i] = 0;
for (int i = 0; i < 100; i++) s[i][i] = 0;
int q = 0;
for (int L = 2; L <= n; L++) {
for (int i = 1; i <= n - L + 1; i++) {
int j = i + L - 1;
m[i][j] = INT_MAX;
for (int k = i; k <= j - 1; k++) {
q = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j];
if (q < m[i][j]) {
m[i][j] = q;
s[i][j] = k;
}
}
}
}
return q;
}
int main()
{
int arr[] = { 40, 20, 30, 10, 30 };
int array_size = sizeof(arr) / sizeof(int);
int n = array_size - 1;
int s[100][100];
int minimum = MatrixChainOrder(s, arr, n);
printf("{ 40, 20, 30, 10, 30 } should result in 26000 : %d\n", minimum);
return 0;
}
同样你可以改变你的Print
函数
void Print(int s[100][100], int i, int j)
{
if (i < 0 || i >= 100 || j < 0 || j >= 100)
{
cout << "array bound error\n";
}
//safely access s[i][j] ...
}