一个二维矩阵存储的世界地图里，每个小方格区域被染上不同的颜色，求此图中的国家总数（上下左右不可斜对角）。



#include <iostream>
#include <stack>
#include <vector>

using namespace std;

int resolution(vector< vector<int> > &A)
{
  //self-defined construct
  struct Point
  {
    int line;
    int col;
    Point(int X, int Y) :
        line(X), col(Y)
    {
    }
    ~Point()
    {
    }
  };
//allocate bitArray for Nodes' visited 0-1 state
  int N = A.size();
  int M = A.at(0).size();
  char *isChecked = new char[(N * M + 8) / 8];
  if (isChecked == NULL)
  {
    return -1;
  }
//initialize all node with all unvisited state.
  for (int i = 0; i < (N * M + 8) / 8; ++i)
  {
    isChecked[i] = 0;
  }

  int currColor = -1;
  stack<Point> currS;// diagram's depth-first traversal algorithm
  int totalCn = 0;

  for (int ii = 0; ii < N; ++ii)
  {
    for (int jj = 0; jj < M; ++jj)
    {
      if ((isChecked[(ii * M + jj) / 8] & (1 << ((ii * M + jj) % 8))) == 0)
      { // if current Node  not yet been visited, set visited.
        isChecked[(ii * M + jj) / 8] |= (1 << ((ii * M + jj) % 8));
        currS.push(Point(ii, jj));// a new 4-branch tree's rootNode
        ++totalCn;
        currColor = A[ii][jj];
        //Start merging all Neighboring Same color Node whose value equals currColor
        while (!currS.empty())
        {
          Point tmp = currS.top();
          currS.pop();
          int i = tmp.line;
          int j = tmp.col;
          //current node's right neighbor if exists
          if ((j + 1 < M)
              && ((isChecked[(i * M + j + 1) / 8]
                  & (1 << ((i * M + j + 1) % 8))) == 0)
              && (A[i][j+1] == currColor))
          {
            isChecked[(i * M + j + 1) / 8] |= (1 << ((i * M + j + 1) % 8));
            currS.push(Point(i, j + 1));
          }
          //current node's down neighbor if exists
          if ((i + 1 < N)
              && ((isChecked[((i+1) * M + j) / 8]
                  & (1 << (((i+1) * M + j) % 8))) == 0)
              && (A[i+1][j] == currColor))
          {
            isChecked[((i+1) * M + j ) / 8] |= (1 << (((i+1) * M + j ) % 8));
            currS.push(Point(i+1, j ));
          }
          //current node's left neighbor if exists
          if ((j - 1 >=0)
              && ((isChecked[(i * M + j - 1) / 8]
                  & (1 << ((i * M + j - 1) % 8))) == 0)
              && (A[i][j-1] == currColor))
          {
            isChecked[(i * M + j - 1) / 8] |= (1 << ((i * M + j - 1) % 8));
            currS.push(Point(i, j - 1));
          }
          //current node's upper neighbor if exists
          if ((i - 1 >=0 )
              && ((isChecked[((i-1) * M + j ) / 8]
                  & (1 << (((i-1) * M + j ) % 8))) == 0)
              && (A[i-1][j] == currColor))
          {
            isChecked[((i-1) * M + j) / 8] |= (1 << (((i-1) * M + j ) % 8));
            currS.push(Point(i-1, j));
          }
        }
      }

    }

  }

  delete[] isChecked;
  isChecked = NULL;

  return totalCn;

}

int main(int argc, char** argv)
{

  return 0;

}