在回溯法的每一步，我們從一個給定的部分解a={a1, a2, ..., ak}開始，嘗試在最后添加元素來擴(kuò)展這個部分解，擴(kuò)展之后，我們必須測試它是否為一個完整解，如果是的話，就輸出這個解；如果不完整，我們必須檢查這個部分解是否仍有可能擴(kuò)展成完整解，如果有可能，遞歸下去；如果沒可能，從a中刪除新加入的最后一個元素，然后嘗試該位置上的其他可能性。

用一個全局變量來控制回溯是否完成，這個變量設(shè)為finished，那么回溯框架如下，可謂是回溯大法之精髓與神器

bool finished = false;

void backTack(int input[], int inputSize, int index, int states[], int stateSize)
{
 int candidates[MAXCANDIDATE];
 int ncandidates;

 if (isSolution(input, inputSize, index) == true)
 {
 processSolution(input, inputSize, index);
 }
 else
 {
 constructCandidate(input, inputSize, index, candidates, &ncandidates);
 for (int i = 0; i < ncandidates; i++)
 {
  input[index] = candidates[i];
  backTack(input, inputSize, index + 1);
  if (finished)
  return;
 }
 }
}

不拘泥于框架的形式，我們可以編寫出如下代碼：

#include <iostream>

using namespace std;

char str[] = "abc";
const int size = 3;

int constructCandidate(bool *flag, int size = 2)
{
 flag[0] = true;
 flag[1] = false;

 return 2;
}

void printCombine(const char *str, bool *flag, int pos, int size)
{
 if (str == NULL || flag == NULL || size <= 0)
 return;
 
 if (pos == size)
 {
 cout << "{ ";
 for (int i = 0; i < size; i++)
 {
  if (flag[i] == true)
  cout << str[i] << " ";
 }
 cout << "}" << endl;
 }
 else
 {
 bool candidates[2];
 int number = constructCandidate(candidates);
 for (int j = 0; j < number; j++)
 {
  flag[pos] = candidates[j];
  printCombine(str, flag, pos + 1, size);
 }
 }
}

void main()
{
 bool *flag = new bool[size];
 if (flag == NULL)
 return;
 printCombine(str, flag, 0, size);
 delete []flag;
}

采用回溯法框架來計算字典序排列：

#include <iostream>

using namespace std;

char str[] = "abc";
const int size = 3;

void constructCandidate(char *input, int inputSize, int index, char *states, char *candidates, int *ncandidates)
{
 if (input == NULL || inputSize <= 0 || index < 0 || candidates == NULL || ncandidates == NULL)
 return;
 
 bool buff[256];
 for (int i = 0; i < 256; i++)
 buff[i] = false;
 int count = 0;
 for (int i = 0; i < index; i++)
 {
 buff[states[i]] = true;
 }
 for (int i = 0; i < inputSize; i++)
 {
 if (buff[input[i]] == false)
  candidates[count++] = input[i];
 }
 *ncandidates = count;
 return;
}

bool isSolution(int index, int inputSize)
{
 if (index == inputSize)
 return true;
 else
 return false;
}

void processSolution(char *input, int inputSize)
{
 if (input == NULL || inputSize <= 0)
 return;

 for (int i = 0; i < inputSize; i++)
 cout << input[i];
 cout << endl;
}

void backTack(char *input, int inputSize, int index, char *states, int stateSize)
{
 if (input == NULL || inputSize <= 0 || index < 0 || states == NULL || stateSize <= 0)
 return;
 
 char candidates[100];
 int ncandidates;
 if (isSolution(index, inputSize) == true)
 {
 processSolution(states, inputSize);
 return;
 }
 else
 {
 constructCandidate(input, inputSize, index, states, candidates, &ncandidates);
 for (int i = 0; i < ncandidates; i++)
 {
  states[index] = candidates[i];
  backTack(input, inputSize, index + 1, states, stateSize);
 }
 }
}

void main()
{
 char *candidates = new char[size];
 if (candidates == NULL)
 return;
 backTack(str, size, 0, candidates, size);
 delete []candidates;
}

對比上述兩種情形，可以發(fā)現(xiàn)唯一的區(qū)別在于全排列對當(dāng)前解向量沒有要求，而字典序?qū)Ξ?dāng)前解向量是有要求的，需要知道當(dāng)前解的狀態(tài)！
八皇后回溯法求解：

#include <iostream>

using namespace std;

int position[8];

void constructCandidate(int *input, int inputSize, int index, int *states, int *candidates, int *ncandidates)
{
 if (input == NULL || inputSize <= 0 || index < 0 || candidates == NULL || ncandidates == NULL)
 return;
 
 *ncandidates = 0;
 bool flag;
 for (int i = 0; i < inputSize; i++)
 {
 flag = true;
 for (int j = 0; j < index; j++)
 {
  if (abs(index - j) == abs(i - states[j]))
  flag = false;
  if (i == states[j])
  flag = false;
 }

 if (flag == true)
 {
  candidates[*ncandidates] = i;
  *ncandidates = *ncandidates + 1;
 }
 }
/*
 cout << "ncandidates = " << *ncandidates << endl;
 system("pause");*/

 return;
}

bool isSolution(int index, int inputSize)
{
 if (index == inputSize)
 return true;
 else
 return false;
}

void processSolution(int &count)
{
 count++;
}

void backTack(int *input, int inputSize, int index, int *states, int stateSize, int &count)
{
 if (input == NULL || inputSize <= 0 || index < 0 || states == NULL || stateSize <= 0)
 return;
 
 int candidates[8];
 int ncandidates;
 if (isSolution(index, inputSize) == true)
 {
 processSolution(count);
 }
 else
 {
 constructCandidate(input, inputSize, index, states, candidates, &ncandidates);
 for (int i = 0; i < ncandidates; i++)
 {
  states[index] = candidates[i];
  backTack(input, inputSize, index + 1, states, stateSize, count);
 }
 }
}

void main()
{
 //初始化棋局
 for (int i = 0; i < 8; i++)
 position[i] = i;

 int states[8];
 int count = 0;
 backTack(position, 8, 0, states, 8, count);
 cout << "count = " << count << endl;
}

希望本文所述對大家C++程序算法設(shè)計的學(xué)習(xí)有所幫助。

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