算法设计与分析实验源代码及运行结果.docx

1、算法设计与分析实验源代码及运行结果1. 分治法二分检索源代码：#include using namespace std;void SelectSort(int*pData, int Countint iTemp;int iPos;for(int i=0; i iTemp = pDatai;iPos = i;for(int j=i+1; j if(pDataj iTemp = pDataj;iPos = j;pDataiPos = pDatai;pDatai = iTemp;void main(int iCount;cout请输入要排序的个数:iCount;int Data2000;cout请输

2、入原始数据: for(int k=0; k coutDataDatak;cout排序之前的数据顺序为: for(int j=0; j cout ; cout cout排序之后的数据顺序为: SelectSort(Data,iCount;for(int i=0; i cout cout 运行结果截图：2.分治法排序源代码：#include #include using namespace std;int divide(int *list, int low, int highint pivotkey = listlow;while (low highwhile (low = pivotkey hi

3、gh-;listlow = listhigh;while (low high & listlow = pivotkey low+;listhigh = listlow;listlow = pivotkey;return low;/一次划分;void sort(int *list, int low, int highif (low highint pivotloc = divide(list, low, high;sort(list, low, pivotloc - 1;sort(list, pivotloc + 1, high;/快速排序算法;int main(int arry100;int

4、i;coutThe arry to be sorted are: for (i = 0; i 100; i+arryi = rand( % 100;cout sort(arry, 0, 99;cout for (i = 0; i 100; i+cout return 0;运行结果截图：1.货郎担问题源代码：#include #include #include #define NUM 4using namespace std;int num = (intpow(2, NUM - 1;class subsetpublic:bool containsNUM - 1; /除了开始结点外，包含哪个城市就

5、把相应的contains设置为1；bool is_in(int ii-;if (containsi return true;else return false;/判断第i个城市是否在子集中，如果在返回true，否则false；;/所有节点城市的子集。void set_ct(subset * s, int begin, int end, int processif (NUM - 2 = processsbegin.containsprocess = false;send.containsprocess = true;elseint middle = (end + begin / 2;int i;

6、for (i = begin; i = middle; i+si.containsprocess = false;set_ct(s, begin, middle, process + 1;for (i = middle + 1; i = 0 & a = bif (0 = b | a = b return 1;elseif (a - b b b = a - b; /组合数的性质；int i, result = 1;for (i = a - b + 1; i = a; i+result *= i;for (i = 2; i = b; i+result /= i;return result;else

7、 return -1;/组合数（a选b）的计算,数据合法返回正确的值，否则返回-1，int get_h(bool *c, int nint i, result = 0;for (i = 0; i NUM - 1; i+if (i != n & ci result += (intpow(2, NUM - 2 - i;return result;/计算table_d表的横坐标horizontal，int main(int *cities = new int *NUM; /存储个城市之间的代价矩阵int *table_d = new int *NUM; /存储动态距离表int *table_p =

8、new int *NUM; /存储动态结点选在表subset *sbst = new subsetnum; /所有城市节点的子集。int i, j, k, temp, horizontal;for (i = 0; i NUM; i+citiesi = new intNUM;table_di = new intnum;table_pi = new intnum;for (j = 0; j citiesij;table_di0 = citiesi0; /初始化距离表的第一列/*cities00 = 32767;cities01 = 3;cities02 = 6;cities03 = 7;citie

9、s10 = 5;cities11 = 32767;cities12 = 2;cities13 = 3;cities20 = 6;cities21 = 4;cities22 = 32767;cities23 = 2;cities30 = 3;cities31 = 7;cities32 = 5;cities33 = 32767;for (i = 1; i NUM; i+table_di0 = citiesi0;*/set_ct(sbst, 0, num - 1, 0;for (j = 1; j num - 1; j+for (i = 1; i NUM; i+if (!sbstj.containsi

10、 - 1table_dij = 32767;for (k = 1; k NUM; k+if (sbstj.containsk - 1horizontal = get_h(sbstj.contains, k - 1; /计算出此时的table_d的横坐标。temp = citiesik + table_dkhorizontal;if (temp 0; i-horizontal = get_h(sbstnum - 1.contains, i - 1;temp = cities0i + table_dihorizontal;if (temp table_d0num - 1table_d0num -

11、1 = temp;table_p0num - 1 = i;coutThe shortest distance is: cout ; temp = table_p0num - 1;horizontal = get_h(sbstnum - 1.contains, temp - 1;for (i = 1; i NUM - 1; i+cout ; temp = table_ptemphorizontal;horizontal = get_h(sbsthorizontal.contains, temp - 1;cout0;return 0;运行结果截图：当各城市距离都相等，比如时：当输入课件中讨论的数据

12、时：当输入上述矩阵按其主对角线对换后的矩阵的数据时：路径刚好跟上面的相反。2.投资问题源代码：#include #include using namespace std;#define N 5 /工程数；#define A 10 /资金总数;int main(int *g = new int*N; /N X A的矩阵，存储对工程i投资j万元的利润；int *table = new int*N + 1; /(N+1 X (A+1矩阵,存入动态决策表；int *cash = new int*N; /N X A的矩阵，存储各工程的投资数int *result = new intN; /记录最终结果，

13、用于输出；int i, j, k, temp;for (i = 0; i 0; i- table0i = 0; /table第一行赋0；coutThe profit of j*10000 yuan invest to project i are: for(i = 0; i N; i+for (j = 0; j A; j+gij = rand( % 5;cout cout /为了方便，随机给利润矩阵g赋值；cout for (i = 1; i = N; i+ /对每一列：for (j = 1; j = A; j+ /对该列的每一个元素tableij = tablei - 1j;cashi - 1j - 1 = 0;for (k = 1; k tableijtableij = temp;cashi - 1j - 1 = k;/记下在有j万元资金的情况下投资前i个工程的最大利润，同时记录投资工程i的资金到cash中temp = A - 1;for (i = N - 1; i = 0; i-resulti = cashitemp;temp -= resulti;/从cash表格中得出最终结果；coutInvestment to each project are: for (i = 0; i N; i+coutProjectcout return 0;运行结果截图：