leecode 解题总结：130. Surrounded Regions

#include <iostream>
#include <stdio.h>
#include <vector>
#include <string>
#include <queue>
using namespace std;
/*
问题:
Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

For example,
X X X X
X O O X
X X O X
X O X X
After running your function, the board should be:

X X X X
X X X X
X X X X
X O X X

输入:
4(行) 4(列)
X X X X
X O O X
X X O X
X O X X
输出:
X X X X
X X X X
X X X X
X O X X

关键
1 分析:此题实际上找到四周都是"X"的"O"，并把这些"O"替换为"X"。
参考这位作者的解法:http://www.cnblogs.com/ganganloveu/p/3755191.html
关键就是从何处开始替换。观察发现在边界上的"O"都不会被替换，先将这部分"O"
转换为"#"，对于最后不是"#"的"O"就全部设置为"X"，然后再把"#"设置为"O"即可
后续的遍历中会对边界的邻居结点上的"O"继续递归操作
*/
struct MyPoint
{
	MyPoint(int row , int col):_row(row),_col(col){}
	MyPoint(){}
	int _row;
	int _col;
};

class Solution {
public:
	//对位置(curRow,curCol)即处于边界的元素上的"O"尝试都设置为"#"
	void bfs(int curRow , int curCol , int row , int col,vector<vector<char>>& board)
	{
		if(curRow < 0 || curRow >= row || curCol < 0 || curCol >= col)
		{
			return;
		}
		//易错，当前元素要设置为"#"
		board.at(curRow).at(curCol) = '#';
		queue<MyPoint> points;
		points.push(MyPoint(curRow , curCol));
		MyPoint point;
		while(!points.empty())
		{
			point = points.front();
			points.pop();
			//向下寻找"O"的元素
			if( (point._row + 1) < row && 'O' == board.at( point._row + 1 ).at(point._col))
			{
				board.at( point._row + 1 ).at(point._col) = '#';
				points.push(MyPoint(point._row + 1, point._col));
			}
			
			//向上
			if( (point._row - 1) >= 0 && 'O' == board.at( point._row - 1 ).at(point._col) )
			{
				board.at(point._row - 1).at(point._col) = '#';
				points.push(MyPoint(point._row - 1 , point._col));
			}

			//向右
			if( (point._col + 1) < col && 'O' == board.at( point._row ).at(point._col + 1) )
			{
				board.at( point._row ).at(point._col + 1) = '#';
				points.push(MyPoint( point._row , point._col + 1));
			}

			//向左
			if( (point._col - 1) >= 0 && 'O' == board.at( point._row ).at(point._col - 1) )
			{
				board.at( point._row ).at(point._col - 1) = '#';
				points.push(MyPoint( point._row , point._col - 1));
			}
		}
	}

    void solve(vector<vector<char>>& board) {
        if(board.empty())
		{
			return;
		}
		int row = board.size();
		int col = board.at(0).size();
		//寻找四个边界上的"O"，然后递归修改为"#"
		for(int i = 0 ; i < row ; i++)
		{
			for(int j = 0 ; j < col ; j++)
			{
				if(0 == i || row - 1 == i || 0 == j || col - 1 == j)
				{
					if('O' == board.at(i).at(j))
					{
						bfs(i , j , row , col , board);
					}
				}
			}
		}
		//将"#"字符替换回原来的"O"
		for(int i = 0 ; i < row ; i++)
		{
			for(int j = 0 ; j < col ; j++)
			{
				if('#' == board.at(i).at(j))
				{
					board.at(i).at(j) = 'O';
				}
				//凡是不是"#"的"O"，都是被围住的，全部修改为"X"
				else if('O' == board.at(i).at(j))
				{
					board.at(i).at(j) = 'X';
				}
			}
		}
    }
};


void print(vector< vector<char> > result)
{
	if(result.empty())
	{
		cout << "no result" << endl;
		return;
	}
	int size = result.size();
	int len = result.at(0).size();
	for(int i = 0 ; i < size ; i++)
	{
		for(int j = 0 ; j < len; j++)
		{
			cout << result.at(i).at(j) << " " ;
		}
		cout << endl;
	}
}

void process()
{
	 vector< vector<char> > board;
	 char value;
	 Solution solution;
	 int row;
	 int col;
	 while(cin >> row >> col )
	 {
		 board.clear();
		 for(int i = 0 ; i < row ; i++)
		 {
			 vector<char> nums;
			 for(int j = 0 ; j < col ; j++)
			 {
				 cin >> value;
				 nums.push_back(value);
			 }
			 board.push_back(nums);
		 }
		 solution.solve(board);
		 print(board);
	 }
}

int main(int argc , char* argv[])
{
	process();
	getchar();
	return 0;
}