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燕山大学《DS&A课程实践》----二级项目子任务

Raow1 2022-03-11 阅读 68
c++

 仅供参考

7-1 ISBN号任务识别工作计划 (10 分)

输入样例:

5 5 
1 2
1 3
2 4 
3 4	
4 5

输出样哩:

在这里给出相应的输出。例如：

1 2 3 4 5 
#include<algorithm>
#include<iostream>
#include<set>
using namespace std;
int main()
{
	int a, b, n, m;
	set<int> g;
	cin >> n >> m;
	for (int i = 0; i < m; i++)
	{
		a = 0; b = 0;
		cin >> a >> b;
		g.insert(a);
		g.insert(b);
	}
	for (set<int>::iterator in=g.begin();in!=g.end();in++)
	{
		cout << *in;
		if (in != g.end())
			cout << " ";
	}
	return 0;
}

7-2 读取包含ISBN号的图像文件的内容 (10 分)

输入样例:

3 3
7 7 8
2 6 9
1 3 5

10 12 14
16 17 19
15 20 11

200 219 207
221 230 218
243 231 229

输出样例:

在这里给出相应的输出。例如：

3 3
7 7 8 
2 6 9 
1 3 5 
5.33

10 12 14 
16 17 19 
15 20 11 
14.89

200 219 207 
221 230 218 
243 231 229 
222.00
#include <iostream>

using namespace std;
int main()
{
    int m, n;
    cin >> m >> n;
    int **arry[3];
    for (int i = 0; i < 3; i++)
        arry[i] = new int* [m];
    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < m; j++)
            arry[i][j] = new int[n];
    }

    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < m; j++)
            for (int k = 0; k < n; k++)
                cin >> arry[i][j][k];
        getchar();
    }
    double sum[3] = { 0 };
    double s;
    for (int i = 0; i < 3; i++)
    {
        s = 0;
        for (int j = 0; j < m; j++)
        {
            for (int k = 0; k < n; k++)
                s += arry[i][j][k];
        }
        sum[i] = s / (m * n);
    }
    cout << m << " " << n << endl;
    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < m; j++)
        {
            for (int k = 0; k < n; k++)
                cout << arry[i][j][k] << " ";
            cout << endl;
        }
        printf("%.2f", sum[i]);
        if (i < 2)
        {
            cout << endl;
            cout << endl;
        }
    }
}

7-3 将24真彩色图转换为灰度图像 (10 分)

输入样例:

3 3 0
3 6 8
4 8 6
3 9 5
1 4 8
5 2 8
2 5 9
2 5 7
7 5 2
3 9 6

输出样例:

1 4 7
4 4 6
2 6 7

#include<iostream>
#include<algorithm>
using namespace std;

//R0.299 + G0.587 + B0.114
void math_0(double*** &a,int n,int m)
{
    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < n; j++)
        {
            for (int k = 0; k < m; k++)
            {
                switch (i)
                {
                case 0:
                    a[i][j][k] *= 0.299;
                    break;
                case 1:
                    a[i][j][k] *= 0.587;
                    break;
                case 2:
                    a[i][j][k] *= 0.114;
                    break;
                default:
                    break;
                }
            }
        }
    }
    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < n; j++)
            for (int k = 0; k < m; k++)
                a[3][j][k] += a[i][j][k];
    }

}
//(R + G + B) / 3
void math_1(double***& a, int n, int m)
{
    for (int i = 0; i < 3; i++)
            for (int j = 0; j < n; j++)
                for (int k = 0; k < m; k++)
                    a[3][j][k] += a[i][j][k];
       
    for (int j = 0; j < n; j++)
        for (int k = 0; k < m; k++)
            a[3][j][k] =a[3][j][k]/ 3;
}
//max(R, G, B)
void math_2(double***& a, int n, int m)
{
    double s[3];
    for (int j = 0; j < n; j++)
    {
        for (int k = 0; k < m; k++)
        {
            s[0] = a[0][j][k];
            s[1] = a[1][j][k];
            s[2] = a[2][j][k];
            sort(s, s + 3);
            a[3][j][k] = s[2];
        }
    }
}
int main()
{
	int n, m;
	char type;
	cin >> n >> m >> type;
    double*** arry = new double** [4];
    for (int i = 0; i < 4; i++)
        arry[i] = new double* [n];
    for (int i = 0; i < 4; i++)
    {
        for (int j = 0; j < n; j++)
            arry[i][j] = new double[m];
    }

    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < n; j++)
            for (int k = 0; k < m; k++)
                cin >> arry[i][j][k];
    }
    for (int j = 0; j < n; j++)
        for (int k = 0; k < m; k++)
            arry[3][j][k] = 0;
	switch (type)
	{
	case '0':
        math_0(arry,n,m);
        for (int j = 0; j < n; j++)
        {
            for (int k = 0; k < m; k++)
            {
                int s = arry[3][j][k];
                cout << s;
                if (k < m - 1)
                    cout << " ";
            }
            if (j < n - 1)
                cout << endl;
        }
		break;
    case '1':
        math_1(arry, n, m);
        for (int j = 0; j < n; j++)
        {
            for (int k = 0; k < m; k++)
            {
                int s = arry[3][j][k];
                cout << s;
                if (k < m - 1)
                    cout << " ";
            }
            if (j < n - 1)
                cout << endl;
        }
        break;
    case '2':
        math_2(arry, n, m);
        for (int j = 0; j < n; j++)
        {
            for (int k = 0; k < m; k++)
            {
                int s = arry[3][j][k];
                cout << s;
                if (k < m - 1)
                    cout << " ";
            }
            if (j < n - 1)
                cout << endl;
        }
        break;
	default:
		break;
	}
	return 0;
}

7-4 做出灰度图的灰度直方图 (10 分)

输入样例:

4 5
1 3 7 9 12 
4 4 8 10 12
4 3 8 9 10
1 4 10 12 8

输出样例:

在这里给出相应的输出。例如：

1:2
3:2
4:4
7:1
8:3
9:2
10:3
12:3

#include<iostream>

using namespace std;
int main()
{
	int n=0, m=0;
	int pin[256] = { 0 };
	cin >> n >> m;
	int a;
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			cin >> a;
			pin[a] += 1;
		}
	}
	for (int i = 0; i < 256; i++)
	{
		if (pin[i] == 0)
			continue;
		cout << i << ":" << pin[i]<<endl;
	}
	return 0;
}

7-5 找最小的值做图像分割 (10 分)

输入样例:

6
3 5 8 6 7 2

输出样例:

2 3 5 6 7 8 
灰度级频数最小的值为：2
#include<iostream>

using namespace std;

void bubbleSort(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        //比较两个相邻的元素   
        for (int j = 0; j < n - i - 1; j++) {
            if (arr[j] > arr[j + 1]) {
                int t = arr[j];
                arr[j] = arr[j + 1];
                arr[j + 1] = t;
            }
        }
    }
}

int main()
{
    int n = 0;
    cin >> n;
    int* a = new int[n];
    for (int i = 0; i < n; i++)
        cin >> a[i];
    bubbleSort(a, n);
    for (int i = 0; i < n; i++)
        cout << a[i] << " ";
    cout << endl;
    cout <<"灰度级频数最小的值为：" <<a[0];
	return 0;
}

7-6 标识每一个数字图元区域 (10 分)

输入样例:

4 6
0 1 1 0 0 1
1 1 1 0 1 1
0 0 0 0 1 1
1 1 1 0 0 1

输出样例:

0 1 1 0 0 2 
1 1 1 0 2 2 
0 0 0 0 2 2 
3 3 3 0 0 2

无（代码有待完善）

7-7 提取每一个数字的最小平行外接矩阵 (10 分)

输入样例:

10 17
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0
0 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 0
0 0 1 1 1 0 0 0 0 0 0 0 1 1 1 0 0
0 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 0
0 0 1 1 1 0 0 0 0 0 0 0 1 1 1 0 0
0 0 0 1 1 1 0 0 0 0 0 1 1 1 0 0 0
0 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 0
0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

输出样例:

(1 1) (8 15)

#include<iostream>

using namespace std;
int main()
{
	int m, n;
	cin >> m >> n;
	int** arry = new int* [m];
	for (int i = 0; i < m; i++)
		arry[i] = new int[n];
	for (int i = 0; i < m; i++)
	{
		for (int j = 0; j < n; j++)
		{
			cin >> arry[i][j];
		}
	}
	int a = 0, b = 0, c = 0, d = 0;
	bool tep = true;
	for (int j = 0; j < n; j++)
	{
		for (int i = 0; i < m; i++)
		{
			if (arry[i][j] == 1 && tep)
			{
				a = j;
				tep = false;
			}
			if (arry[i][j] == 1)
				b = j;
		}
	}
	tep = true;
	for (int i = 0; i < m; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (arry[i][j] == 1 && tep)
			{
				c = i;
				tep = false;
			}
			if (arry[i][j] == 1)
				d = i;
		}
	}
	printf("(%d %d) (%d %d)", c, a, d, b);
	return 0;
}

7-8 数字图像尺度归一化 (10 分)

输入样例:

2 2 7 7
1 0
0 1

输出样例:

1 1 1 1 0 0 0
1 1 1 1 0 0 0
1 1 1 1 0 0 0
1 1 1 1 0 0 0
0 0 0 0 1 1 1
0 0 0 0 1 1 1
0 0 0 0 1 1 1

#include<iostream>
using namespace std;
int main()
{
	int m, n, p, q;
	cin >> m >> n >> p >> q;
	int** arry = new int * [m];
	int** arry1 = new int* [p];
	for (int i = 0; i < m; i++)
		arry[i] = new int[n];
	for (int i = 0; i < p; i++)
		arry1[i] = new int[q];
	for (int i = 0; i < m; i++)
		for (int j = 0; j < n; j++)
			cin >> arry[i][j];
	int a, b;
	for (int i = 0; i < p; i++)
	{
		for (int j = 0; j < q; j++)
		{
			a = i * m / p;
			b = j * n / q;
			arry1[i][j] = arry[a][b];
			cout << arry1[i][j];
			if (j < q - 1)
				cout << " ";
		}
		if (i < p - 1)
			cout << endl;
	}
		
	return 0;
}

7-9 中值滤波 (10 分)

输入样例:

4 4
1 5 6 7
3 4 2 8
9 6 3 2
1 5 8 6

输出样例:

1 5 6 7
3 4 5 8
9 4 5 2
1 5 8 6

#include<iostream>
#include<algorithm>
using namespace std;

int paixu(int** a)
{
	int* b = new int[100];
	int s = 0;
	int index = 0;
	for (int i = 0; i < 3; i++)
	{
		for (int j = 0; j < 3; j++)
		{
			b[index] = a[i][j];
			index++;
		}
	}
	sort(b, b + 9);
	s = b[4];
	return s;
}
int main()
{
	int n, m;
	cin >> n >> m;
	int** arry = new int* [n];//用于存放输入的矩阵
	int** midd = new int* [n];//用于存放矩阵（除边缘外）改变后的数值
	for (int i = 0; i < n; i++)
	{
		arry[i] = new int[m];
		midd[i] = new int[m];
	}

	for (int i = 0; i < n ; i++)
	{
		for (int j = 0; j < m ; j++)
		{
			cin >> arry[i][j];
		}
	}
	int** tep = new int* [3];
	for (int i = 0; i < 3; i++)
		tep[i] = new int[3];
	
	for (int i = 1; i < n - 1; i++)
	{
		for (int j = 1; j < m - 1; j++)
		{
			tep[1][1] = arry[i][j];
			tep[0][1] = arry[i-1][j];
			tep[1][0] = arry[i][j-1];
			tep[0][0] = arry[i-1][j-1];
			tep[0][2] = arry[i-1][j+1];
			tep[1][2] = arry[i][j+1];
			tep[2][0] = arry[i+1][j-1];
			tep[2][1] = arry[i+1][j];
			tep[2][2] = arry[i+1][j+1];
			midd[i][j] = paixu(tep);
		}
	}
	for (int i = 1; i < n - 1; i++)
	{
		for (int j = 1; j < m - 1; j++)
		{
			arry[i][j] = midd[i][j];
		}
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			cout << arry[i][j];
			if (j < m - 1)
				cout << " ";
		}
		if (i < n - 1)
			cout << endl;
	}
	return 0;
}

7-10 计算正确率和准确率 (10 分)

输入样例:

3
9787111478188 9787111470108
927086257902X 927086257902X
8927518936148 093851895614X

输出样例:

0.33 0.82
#include<iostream>
#include<string>
using namespace std;
int main()
{
	int n;
	cin >> n;
	string** str = new string* [n];
	for (int i = 0; i < n; i++)
		str[i] = new string[2];
	for (int i = 0; i < n; i++)
		for (int j = 0; j < 2; j++)
			cin >> str[i][j];
	double flag = 0, s,t=0, num = 0;
	for (int i = 0; i < n; i++)
	{
		s = 0;
		t += str[i][0].size();
		for (int j = 0; j < str[i][0].size(); j++)
		{
			if (str[i][1][j] == str[i][0][j])
				s++;
		}
		num += s;
		if (s == str[i][0].size())
			flag++;
	}
	printf("%.2f %.2f", flag / n, num / t);
	return 0;
}

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