1. 按字典序生成组合
// 生成字典序
void dict_gen(int *array, int len)
{
int s = 1<<len; // 要进行 2^n 次循环
for(int i=0;i<s; i++)
{
// 递增序列刚好对应了字典序的顺序
print_dict(array, len, i);
cout<<endl;
}
}
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//显示字典序
void print_dict(int *array, int len, int key)
{
if(len == 0 || array == NULL)
return;
int curr = 1;
std::bitset<8> aout(key);
cout<<aout<<": ";
for(int i=0;i<len;i++)
{
// curr为当前下标对应的字典序,如下标为3对应的字典序为:1000 0000
curr = 1 << i;
// key为一个综合的字典序,如 key = 0100 1100
// 如果 key在curr位为1的话,则会打印出来
if(key & curr)
{
cout//<<setw(4)
<<array[i]<<',';
}
}
}
#include <iostream>
#include <bitset>
#include <stdlib.h>
using namespace std;
using std::bitset;
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#define BIT_WIDTH 8
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// gray 码生成器
void gray(int *array, int len);
// 用于找到最后一个1的位置,如果找不到,则返回-1
int findlast1pos(int value);
// 用于打印gray码
void print_dict(int *array, int len, int key);
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void gray(int *array, int len)
{
// 最后一个数,为: 10
0
int target = (1<<len - 1);
// 循环次数,为:2^n
int total = (1<<len);
// gray序的字典序,默认从0开始
int dict = 0;
// 相加因子,用于将其中的某一位置0或置1
int factor = 0;
// 表示1的个数是否为奇数,0为否
int oddFlag = 0;
int v;
for(int i=0;i<total;i++)
{
print_dict(array, len, dict);
// 如果是最后一个数,则退出
if(dict == target)
break;
// 如果有偶数个1
if(oddFlag == 0)
{
if(dict & 0x01) //如果最后一位为 1,则置 0
dict--;
else
dict = dict | 0x01; // 否则置1
oddFlag = 1;
}
else // 如果是奇数个1
{
v = findlast1pos(dict); // 找到最后一个1的位置
if(v == -1)
break;
// 将那一位反向(0变1,1变0)
factor = 1<<(v+1);
if(dict & factor)
dict = dict & (~factor); // 将第v位置0
else
dict |= factor; // 将第v位置1
oddFlag = 0;
}
cout<<endl;
}
}
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// 算法是:将当前值反复右移并计数,直到最低位为1
int findlast1pos(int value)
{
int ret = -1;
while(value != 0)
{
ret++;
if(value & 0x01)
break;
value >>= 1;
}
return ret;
}
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//显示字典序
void print_dict(int *array, int len, int key)
{
if(len == 0 || array == NULL)
return;
int curr = 1;
std::bitset<BIT_WIDTH> aout(key);
cout<<aout<<": ";
for(int i=0;i<len;i++)
{
// curr为当前下标对应的字典序,如下标为3对应的字典序为:000 0100
curr = 1 << i;
// key为一个综合的字典序,如 key = 0100 1100
// 如果 key在curr位为1的话,则会打印出来
if(key & curr)
{
cout//<<setw(4)
<<array[i]<<',';
}
}
}
#define NUM_1 '1'
#define NUM_0 '0'
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/**
* 生成r-组合的类
*/
class RCombination
{
private:
char *dict;
int *array;
int size;
int r;
private:
void initDict();
void findActive1(int &activeIndex, int &count1);
void adjustDict(int activeIndex, int count1);
void printDict();
public:
RCombination();
~RCombination();
void genRComb(int *array, int size, int r);
};
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RCombination::RCombination()
{
size = 0;
r = 0;
dict = NULL;
}
RCombination::~RCombination()
{
if(dict != NULL)
delete []dict;
}
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void RCombination::genRComb(int *array, int size, int r)
{
if(array == NULL || size == 0 || r == 0)
return;
if(r>size)
return;
this->array = array;
this->size = size;
this->r = r;
initDict();
int count1 = 0;
int activeInex = -1;
do{
cout<<dict<<": ";
printDict();
cout<<endl;
findActive1(activeInex, count1);
adjustDict(activeInex, count1);
}while(activeInex >= 0);
}
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// 初始化基二的字典序
void RCombination::initDict()
{
if(dict != NULL)
delete []dict;
if(size == 0)
{
dict == NULL;
return;
}
dict = new char[size + 1];
// 使用 1
10
0初始化字典序
for(int i=0;i<size;i++)
{
if(i<r) dict[i] = NUM_1;
else dict[i] = NUM_0;
}
dict[size] = '\0';
}
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// 从后往前找到第一个活动的 1
// 活动的 1 定义如下:只要 1 后面有 0 (并不一定是紧接)
void RCombination::findActive1(int &activeIndex, int &count1)
{
int meet0 = 0;
count1 = 0;
activeIndex = -1;
for(int i=size-1;i>=0;i--)
{
if(dict[i] == NUM_0)
{
meet0 = 1;
}
else
{
count1++;
if(meet0 == 1)
{
activeIndex = i;
break;
}
}
}
}
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// 设i是从最后面开始第1个活动的 1,将 a1, a2,
, ai, ai+1,
变成
// a1, a2,
ai-1, ai+1, ai+2,
void RCombination::adjustDict(int activeIndex, int count1)
{
if(activeIndex >= 0)
{
dict[activeIndex++] = NUM_0;
for(;activeIndex < size; activeIndex++, count1--)
{
if(count1 <= 0)
dict[activeIndex] = NUM_0;
else
dict[activeIndex] = NUM_1;
}
}
}
void RCombination::printDict()
{
if(array == NULL)
return;
for(int i=0;i<size;i++)
{
if(dict[i] == NUM_1)
cout<<array[i]<<',';
}
}
终于贴完了……
1.1.算法的描述
(1)从0...0开始,当字典序为1...1时,退出
(2)求出使得ai=0的最小整数i
(3)用1代替ai并用0代替ai-1, ai-1, ..., a0
2. 按格雷码序生成组合
2.1.算法描述
(1)从0...0开始,到10...0结束。
(2)设t代表代码中1的个数,若t为偶数,将最后一位变换(1变0,0变1)。
(3)若t为奇数,则找到最后一个1,然后把i+1位变换。
2.2.算法分析
通过算法分析,可以得出以下的结论
1. t的奇偶性是交替出现的,最开始为偶数。
2. 在C语言中,可以通过与一个数,将某一位变成0,通过或一个数将某变成1.
3. 可以通过循环右移,然后检查最后一位是否为1来找到最后一个1.
4. 通过判断最后一位是否为(二进制)1来判断是奇数还是偶数。
3. 生成r-组合
3.1.算法描述
(1)算法从1..10...0开始,到0...01...1结束。
(2)确定最大的整数k,使得ak +1<= n且ak + 1不是a1, a2,..., ar。
(3)打印新的r-组合a1...ak-1, ak+1,...,ak+r-k+1。
3.2.算法分析
1. 若将a1, a2, ..,ak也用字典序表示的话,生成r-组合就想法于将r个1和k-r个0进行排列。由于整个过程中,1的数目和0 的数目都是不变的,可以用一个长为k的字符串来表示,字符串中有r个1和k-r个0。r个1和k-r个0的每一个排列都对应一个组合。
2. 引入活动的1这个概念。ai=1为活动的1仅当存在aj=0且j>i。直接的说法就是仅当这个1的后面有0。
按字典序生成组合的实现
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按格雷码序生成组合的实现
<代码>
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生成r-组合的实现
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终于贴完了……