求集合子集问题

//
src是源数据集合,currentIndex是在源集合里的当前下标,length为源集合的大小,dest是结果集合,num是结果集合的元素个数,初始化时,结
果集合要和源集合的个数相等.

void match( int * src, int currentIndex, int length, int * dest,
int num)
![](http://images.csdn.net/syntaxhighlighting/OutliningIndicators/ExpandedBloc
kStart.gif) ![](http://images.csdn.net/syntaxhighlighting/OutliningIndicators/
ContractedBlock.gif) … {

if (currentIndex == length)

return ;


dest[num ++ ] = src[currentIndex ++ ];//采用当前元素

for ( int i = 0 ; i < num; i ++ )
![](http://images.csdn.net/syntaxhighlighting/OutliningIndicators/ExpandedSubB
lockStart.gif) ![](http://images.csdn.net/syntaxhighlighting/OutliningIndicato
rs/ContractedSubBlock.gif) … {

cout << dest[i] << " " ;
![](http://images.csdn.net/syntaxhighlighting/OutliningIndicators/ExpandedSubB
lockEnd.gif) }

cout << endl;

match(src,currentIndex,length,dest,num);//选用当前元素进行递归

match(src,currentIndex,length,dest,num - 1 );//去除当前元素再进行递归
![](http://images.csdn.net/syntaxhighlighting/OutliningIndicators/ExpandedBloc
kEnd.gif) }

int src[] = {1,2,3};
int* dest = new int[3];打印的结果是：

1
1 2
1 2 3
1 3
2
2 3
3

所以这个算法可以求背包问，加和问题等。凡是与子集合相关的问题，全部可以用这个方法解决。