生成{1,2,..n}的字典序r-组合算法

组合的概念不用多解释。计算组合数有特定的公式。

当然某些应用可能要求我们枚举组合中所有的可能情况，这时套公式算数就无能为力了。

如：集合{1,2,3,4}的字典序2-组合为{1,2}{1,3}{1,4}{2,3}{2,4}{3,4}，组合数为6。

计算机生成组合各种情况有多种方式，主要的思路是，首先要凑够r-组合，即数量要够，等于r；其次，不能重复，可以依次列举包含第一个元素的r-组合，列举完毕后，列举包含第二个元素，但不包含第一个元素的r-组合，以此类推。

我用递归的方法实现了一个基于上述思路的生成组合算法，求解{1,2....8}的4-组合。

#define ARRAYSIZE 8#define PICKNUM 4int gl_i = 0;//统计组合数//对于输入inputArray[i...ARRAYSIZE-1]，取k个元素的所有组合，逆序存入outputArray中void Recur_Combination(int inputArray[], int i, int k, int outputArray[]);int main(int argc, char* argv[]){int inputArray[ARRAYSIZE] = {1, 2, 3, 4, 5, 6, 7, 8};int outputArray[PICKNUM];//递归方法求组合Recur_Combination(inputArray, 0, PICKNUM, outputArray);cout << endl << gl_i << endl;return 0;}void Recur_Combination(int inputArray[], int i, int k, int outputArray[]){if (k == 0){for (int j = 0; j < PICKNUM; j++){cout << outputArray[j] << " ";}gl_i++;cout << endl;}else if (i + k - 1 <= ARRAYSIZE - 1){outputArray[k - 1] = inputArray[i];//选取inputArray中第i个元素放入组合，则在剩下的inputArray[i+1..ARRAYSIZE-1]中，选取k-1个Recur_Combination(inputArray, i+1, k-1, outputArray);//若没有选取第i个元素，则在剩下的inputArray[i+1..ARRAYSIZE-1]中，选取k个Recur_Combination(inputArray, i+1, k, outputArray);}}

 

上述算法生成组合序列的效率并不是很高。不适合用来生成大规模组合序列。下面介绍一种较高效的生成方法。

设集合为{1,2,....n}，需生成该集合的r-组合。核心思想是：对于当前已经得到的r-组合{a1,a2,....ar}，找到最大的k，使得ak<n，并且ak+1没有出现在r-组合{a1,a2,....ar}中。程序实现如下：

//////////////////////////////////////////////////////////////////////////// This program is designed to create r- combination in alphabet order.// The main idea in this program is to find the max intergar k so that // ak < n and ak + 1 doesn't appear in a1a2a3....ar.//////////////////////////////////////////////////////////////////////////void GetCombination(int iTotal, int iPickOut);bool IsPickOutOver(int *pPickOutArrany, int iPickOut, int iTotal);bool IsNumInPickOutArray(int num, int *pPickOutArrany, int iPickOut);void PrintPickOutArray(int *pPickOutArrany, int iPickOut);int gl_i = 0;//count the number of differnt combinationsint main(int argc, char* argv[]){int i, j;printf("Please enter the number of the elments:/n");scanf("%d", &i);printf("Please enter the number you want to pick:/n");scanf("%d", &j);GetCombination(i, j);printf("The number of different combinations is: %d./n", gl_i);return 0;}void GetCombination(int iTotal, int iPickOut){// this program only create the PickOutArray[1..iPickOut] to store the picked index of the CollectionArray[1...iTotal](CA).// iPickOut <= iTotal. This Program doesn't have CA. Just default set CA = 1..iTotal.// You can get the real elements in the CollectionArray by index.int *pPickOutArrany = (int *)malloc((iPickOut + 1) * sizeof(int *));//0th is not usedint i;for (i = 1; i <= iPickOut; i++){pPickOutArrany[i] = i;}int k;//used to find the max number that a(k-1) doesn't need to changePrintPickOutArray(pPickOutArrany, iPickOut);while (!IsPickOutOver(pPickOutArrany, iPickOut, iTotal)){for (k = iPickOut; k >= 1; k--){//pPickOutArrany[k] must less than iTotal. if pPickOutArray[k]=iTotal, k must be equal to iPickOut,//k need to move leftwards.if (pPickOutArrany[k] < iTotal){//find the max k so that pPickOutArrany[k]+1 is not in pPickOutArray. if (!IsNumInPickOutArray(pPickOutArrany[k]+1, pPickOutArrany, iPickOut)){//update the pPickOutArrany.pPickOutArrany[1..k-1] does't need to changepPickOutArrany[k] += 1;for (i = k + 1; i <= iTotal; i++){pPickOutArrany[i] = pPickOutArrany[i-1] + 1; }PrintPickOutArray(pPickOutArrany, iPickOut);break;}}}}}bool IsPickOutOver(int *pPickOutArrany, int iPickOut, int iTotal)//pick out over?{bool b = true;int i;for (i = 1; i <= iPickOut; i++){if (pPickOutArrany[i] != iTotal - iPickOut + i){b = false;break;}}return b;}bool IsNumInPickOutArray(int num, int *pPickOutArrany, int iPickOut){int i;for (i = 1; i <= iPickOut; i++){if (pPickOutArrany[i] == num){return true;}}return false;}void PrintPickOutArray(int *pPickOutArrany, int iPickOut){int i;for (i = 1; i <= iPickOut; i++){printf("%d ", pPickOutArrany[i]);}printf("/n");++gl_i;}