算法实验2.2、2.3

2.2主要内容

比较快速排序，归并排序以及堆排序算法的时间效率。了解影响算法执行时间的主要因素以及如何降低算法的执行时间。

#include<iostream>
using namespace std;
#include<stdio.h>
#include<malloc.h>
#include<stdlib.h>
#include<time.h>
#include <sys/timeb.h>
#include"string.h"
#pragma warning(disable : 4996)typedef struct RedType {int key;
}RedType;
typedef struct SqList {RedType* r;int Length;
}SqList;SqList* CreateRandomSqList(int sqListLen) {SqList* sq;int i;sq = (SqList*)malloc(sizeof(SqList));//输入序列长度sq->Length = sqListLen;sq->r = (RedType*)malloc(sizeof(RedType) * (sq->Length + 1)); //给序列的每个元素分配空间srand((unsigned)(time(NULL)));for (i = 1; i <= sq->Length; i++) {sq->r[i].key = int(rand());}return sq;  //返回序列起始地址
}
//创建一个与csp一样的存储空间
SqList* CopyRandomSqList(SqList* csp) {SqList* sq;int i;sq = (SqList*)malloc(sizeof(SqList));//输入序列长度sq->Length = csp->Length;sq->r = (RedType*)malloc(sizeof(RedType) * (sq->Length + 1)); //给序列的每个元素分配空间for (i = 1; i <= sq->Length; i++) {sq->r[i].key = csp->r[i].key;}return sq;  //返回序列起始地址
}
void WritetoFile(int num, int sortTime[], FILE* fp) {char ch[20];for (int i = 0; i < num; i++) {_itoa_s(sortTime[i], ch, 10);strcat_s(ch, ",");fwrite(ch, sizeof(char), strlen(ch), fp);}fwrite("\n", sizeof(char), 1, fp);
}//快速排序
void QuickSort(RedType q[], int l, int r) {if (l >= r)return;int i = l - 1, j = r + 1, x = q[l + r >> 1].key;while (i < j) {do i++; while (q[i].key < x);do j--; while (q[j].key > x);if (i < j)swap(q[i].key, q[j].key);}QuickSort(q, l, j);QuickSort(q, j + 1, r);
}
//堆排序
void HeapAdjust(RedType* SR, int s, int m)//一次筛选的过程
{int rc, j;rc = SR[s].key;for (j = 2 * s; j <= m; j = j * 2)//通过循环沿较大的孩子结点向下筛选{if (j < m&& SR[j].key < SR[j + 1].key) j++;//j为较大的记录的下标if (rc > SR[j].key) break;SR[s] = SR[j]; s = j;}SR[s].key = rc;//插入
}
void HeapSort(RedType* SR, int n)
{int temp, i, j;for (i = n / 2; i > 0; i--)//通过循环初始化顶堆{HeapAdjust(SR, i, n);}for (i = n; i > 0; i--){temp = SR[1].key;SR[1].key = SR[i].key;SR[i].key = temp;//将堆顶记录与未排序的最后一个记录交换HeapAdjust(SR, 1, i - 1);//重新调整为顶堆}
}
//归并排序
void aMerge(RedType* SR, int i, int m, int n) {int j, k;for (j = m + 1; j <= n; j++) {if (SR[j].key <= SR[j - 1].key) {int temp = SR[i].key;for (k = j - 1; temp < SR[k].key && k >= i; --k)SR[k + 1].key = SR[k].key;SR[k + 1].key = temp;}}return;
}
void aMSort(RedType* SR, int s, int t)
{if (s < t) {int mid = (s + t) / 2;aMSort(SR, s, mid);aMSort(SR, mid + 1, t);aMerge(SR, s, mid, t);}return;
}
void MergeSort(SqList* L) {aMSort(L->r, 1, L->Length);return;
}int main() {SqList* L, * L1, * L2;struct __timeb64 stime, etime;long int rmtime, rstime;char ch[20];int quickSortTime[105], mergeSortTime[105], heapSortTime[105];FILE* fp;fp = fopen("Curv.csv", "w");for (int i = 10000; i <= 1000000; i = i + 10000) {_itoa_s(i, ch, 10);strcat_s(ch, ",");fwrite(ch, sizeof(char), strlen(ch), fp);}fwrite("\n", sizeof(char), 1, fp);int k = 0;for (int i = 10000; i <= 1000000; i = i + 10000) {L = CreateRandomSqList(i);L1 = CopyRandomSqList(L);L2 = CopyRandomSqList(L);//归并排序_ftime64_s(&stime);MergeSort(L);_ftime64_s(&etime);free(L->r);free(L);rstime = etime.time - stime.time;rmtime = rstime * 1000;rmtime += etime.millitm - stime.millitm;mergeSortTime[k] = rmtime;//快速排序_ftime64_s(&stime);QuickSort(L1->r, 0, L1->Length - 1);_ftime64_s(&etime);free(L1->r);free(L1);rstime = etime.time - stime.time;rmtime = rstime * 1000;rmtime += etime.millitm - stime.millitm;quickSortTime[k] = rmtime;//堆排序_ftime64_s(&stime);HeapSort(L2->r, L2->Length - 1);_ftime64_s(&etime);free(L2->r);free(L2);rstime = etime.time - stime.time;rmtime = rstime * 1000;rmtime += etime.millitm - stime.millitm;heapSortTime[k] = rmtime;k++;}WritetoFile(100, mergeSortTime, fp);WritetoFile(100, quickSortTime, fp);WritetoFile(100, heapSortTime, fp);fclose(fp);return 1;
}

2.3主要内容

学习分析递归程序结构的时间复杂度和影响算法运行时间的因素。

#include<iostream>
#include<stdio.h>
#include<malloc.h>
#include<stdlib.h>
#include<time.h>
#include <sys/timeb.h>
#include"string.h"
#pragma warning(disable : 4996)
#define M 4
#define N 15
using namespace std;class Color {friend int mColoring(int, int, int**);
public:bool Ok(int k);void Backtrack(int k);int n, m, ** a, * x;long sum;
};
long long  cifang(int n,int m)
{int count = 0;long long c = 1;for (int i = 0; i < n; i++){c = c*(m/2);//防止时间复杂度太大了，进行缩小count++;}return c;
}
void WritetoFile(int num, int sortTime[], FILE* fp) {char ch[20];for (int i = 0; i < num; i++) {_itoa_s(sortTime[i], ch, 10);strcat_s(ch, ",");fwrite(ch, sizeof(char), strlen(ch), fp);}fwrite("\n", sizeof(char), 1, fp);
}
int mColoring(int, int, int**);
int main() {int n, m, ** a, sum; long long timecom{ 0 };//cout << "Please input the number of colors:";//cin >> m;struct __timeb64 stime, etime;long int rmtime, rstime;char ch[20];int paintTime[105]{0};FILE* fp;fp = fopen("Curv.csv", "w");for (int i = 5; i <= N; i = i + 1) {_itoa_s(i, ch, 10);strcat_s(ch, ",");fwrite(ch, sizeof(char), strlen(ch), fp);}fwrite("\n", sizeof(char), 1, fp);int k{ 0 };/*for (int i = 5; i <= N; i = i + 1) {//cout << "Please input the number of nodes:";n=i;a = new int* [n + 1];for (int i = 0; i <= n; i++) {a[i] = new int[n + 1];}//cout << "Please input the ralation of nodes:1-->Connected,0-->Not connected";for (int i = 1; i <= n; i++) {for (int j = i + 1; j <= n; j++) {//cout << "please input the ralation of" << i << "and" << j << ":";//cin >> a[i][j];//a[i][j] = 1;//完全图,最外面的i循环换成 i <= M，注意要改四处a[i][j] = 0;//一条边都没有，最外面的i循环换成 i <= Na[j][i] = a[i][j];}}_ftime64_s(&stime);sum = mColoring(n, M, a);_ftime64_s(&etime);rstime = etime.time - stime.time;rmtime = rstime * 1000;rmtime += etime.millitm - stime.millitm;paintTime[k] = rmtime;k++;delete[]a;}WritetoFile(11, paintTime, fp);for (int i = 5; i <= N; i = i + 1){timecom = i * cifang(i,M);_itoa_s(timecom, ch, 10);strcat_s(ch, ",");fwrite(ch, sizeof(char), strlen(ch), fp);}fwrite("\n", sizeof(char), 1, fp);*/k = 0;n = 10;for (int j = 5; j <= N; j = j + 1) {//cout << "Please input the number of nodes:";m = j;a = new int* [n + 1];for (int i = 0; i <= n; i++) {a[i] = new int[n + 1];}//cout << "Please input the ralation of nodes:1-->Connected,0-->Not connected";for (int i = 1; i <= n; i++) {for (int j = i + 1; j <= n; j++) {//cout << "please input the ralation of" << i << "and" << j << ":";//cin >> a[i][j];//a[i][j] = 1;//完全图，注意要改四处a[i][j] = 0;//一条边都没有a[j][i] = a[i][j];}}_ftime64_s(&stime);sum = mColoring(n, m, a);_ftime64_s(&etime);rstime = etime.time - stime.time;rmtime = rstime * 1000;rmtime += etime.millitm - stime.millitm;paintTime[k] = rmtime;k++;delete[]a;}WritetoFile(11, paintTime, fp);for (int i = 5; i <= N; i = i + 1){timecom = n * cifang(n,i);_itoa_s(timecom, ch, 10);strcat_s(ch, ",");fwrite(ch, sizeof(char), strlen(ch), fp);}fwrite("\n", sizeof(char), 1, fp);fclose(fp);//cout << sum << endl;return 1;
}
bool Color::Ok(int k)
{for (int j = 1; j <= n; j++)if ((a[k][j] == 1 && x[j] == x[k]))return false;return true;
}
void Color::Backtrack(int t) {if (t > n) {sum++;/*for (int i = 1; i <= n; i++)cout << x[i] << " ";cout << endl;*/}else {for (int i = 1; i <= m; i++) {x[t] = i;if (Ok(t))Backtrack(t + 1);x[t] = 0;}}
}
int mColoring(int n, int m, int** a) {Color X;X.n = n;X.m = m;X.a = a;X.sum = 0;int* p = new int[n + 1];for (int i = 0; i <= n; i++)p[i] = 0;X.x = p;X.Backtrack(1);delete[]p;return X.sum;
}