1020. 飞地的数量

这题很简单，也是普通的遍历。在四个边上把所有的1遍历一遍，就知道了所有能到边上的点，和grid比较一下就能算出有多少点能走出这个矩阵。再统计一下一共有多少个点，做个减法就OK了。

#include<stdc++.h>

using namespace std;
int countt = 0;

void dfs(std::vector<vector<int>>&grid, int x, int y)
{
	if(x<0 || x>=(int)grid.size() || y<0 || y>=(int)grid[0].size())
		return;
	if(grid[x][y]==1)
	{
		countt++;
		grid[x][y]=2;
		dfs(grid, x+1, y);
		dfs(grid, x, y+1);
		dfs(grid, x-1, y);
		dfs(grid, x, y-1);
	}

}

int numEnclaves(std::vector<vector<int>>& grid) {

	if(grid.empty())
		return 0;

	int onecount = 0;
	for(int i=0; i<(int)grid.size(); i++)
		for(int j=0; j<(int)grid[0].size(); j++)
			if(grid[i][j]==1)
				onecount++;
	for(int i=0; i<(int)grid.size(); i++)
	{
		if(grid[i][0]==1)
			dfs(grid,i,0);
		if(grid[i][grid[0].size()-1]==1)
			dfs(grid,i,grid[0].size()-1);
	}

	for(int j=0; j<(int)grid[0].size(); j++)
	{
		if(grid[0][j]==1)
			dfs(grid,0,j);
		if(grid[grid.size()-1][j]==1)
			dfs(grid,grid.size()-1,j);
	}
	return onecount - countt;
}

int main()
{
	int x[4][4] =  {
   {0,1,1,0},{0,0,1,0},{0,0,1,0},{0,0,0,0}};
	vector<vector<int>> a;
	for(int i=0; i<4; i++)
	{
		vector<int>b;
		for(int j=0; j<4; j++)
		{
			b.push_back(x[i][j]);
		}
		a.push_back(b);
	}
	cout<<numEnclaves(a)<<endl;
	return 0;
}