代码随想录算法训练营day52

1.孤岛的总面积

1.1 题目

1.2 题解

#include <iostream>
#include <vector>
#include <queue>
#include <list>
using namespace std;


int dir[4][2] = { 0,1,1,0,-1,0,0,-1 };
int count;

void dfs(vector<vector<int>>& grid, int x, int y)
{
	grid[x][y] = 0;
	count++;
	for (int i = 0; i < 4; i++)
	{
		int nextx = x + dir[i][0];
		int nexty = y + dir[i][1];

		//超过边界不考虑
		if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size())continue;
		if (grid[nextx][nexty] == 0)continue;

		dfs(grid, nextx, nexty);
	}
	
}
int main()
{
	
	int n, m;
	cin >> n >> m;
	vector<vector<int>> grid(n, vector<int>(m, 0));
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			cin >> grid[i][j];
		}
	}
	for (int i = 0; i < n; i++)
	{
		if (grid[i][0] == 1)dfs(grid, i, 0);
		if (grid[i][m - 1] == 1)dfs(grid, i, m - 1);
	}
	for (int i = 0; i < m; i++)
	{
		if (grid[0][i] == 1)dfs(grid, 0, i);
		if (grid[n - 1][i] == 1)dfs(grid, n - 1, i);
	}
	count = 0;
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			if (grid[i][j] == 1)dfs(grid, i, j);
		}
	}
	cout << count << endl;
}

2.沉没孤岛

2.1 题目

102. 沉没孤岛

2.2 题解

#include <iostream>
#include <vector>
#include <queue>
#include <list>
using namespace std;


int dir[4][2] = { 0,1,1,0,-1,0,0,-1 };


void dfs(vector<vector<int>>& grid, int x, int y)
{
	grid[x][y] = 2;
	for (int i = 0; i < 4; i++)
	{
		int nextx = x + dir[i][0];
		int nexty = y + dir[i][1];

		//超过边界不考虑
		if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size())continue;
		if (grid[nextx][nexty] == 0||grid[nextx][nexty]==2)continue;

		dfs(grid, nextx, nexty);
	}
	
}
int main()
{
	
	int n, m;
	cin >> n >> m;
	vector<vector<int>> grid(n, vector<int>(m, 0));
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			cin >> grid[i][j];
		}
	}
	for (int i = 0; i < n; i++)
	{
		if (grid[i][0] == 1)dfs(grid, i, 0);
		if (grid[i][m - 1] == 1)dfs(grid, i, m - 1);
	}
	for (int i = 0; i < m; i++)
	{
		if (grid[0][i] == 1)dfs(grid, 0, i);
		if (grid[n - 1][i] == 1)dfs(grid, n - 1, i);
	}

	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			if (grid[i][j] == 1)grid[i][j]=0;
			if (grid[i][j] == 2)grid[i][j] = 1;
		}
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			cout << grid[i][j] << " ";
		}
		cout << endl;
	}

}

3.水流问题

3.1 题目

103. 水流问题

3.2 题解

#include <iostream>
#include <vector>
#include <queue>
#include <list>
using namespace std;


int dir[4][2] = { 0,1,1,0,-1,0,0,-1 };

void dfs(const vector<vector<int>>& grid, vector<vector<bool>>& border, int x, int y)
{
	if (border[x][y])return;
	border[x][y] = true;
	for (int i = 0; i < 4; i++)
	{
		int curx = x + dir[i][0];
		int cury = y + dir[i][1];
		//如果越界跳过
		if (curx < 0 || curx >= grid.size() || cury < 0 || cury >= grid[0].size())continue;
		//逆序
		if (grid[curx][cury] < grid[x][y])continue;
		dfs(grid,border,curx,cury);
	}
}

int main()
{
	int n, m;
	cin >> n >> m;
	vector<vector<int>> grid(n, vector<int>(m, 0));
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			cin >> grid[i][j];
		}
	}
	//标记第一组边界上的节点出发，可以遍历的节点
	vector<vector<bool>> firstborder(n, vector<bool>(m, false));
	//标记第二组边界上的节点出发，可以遍历的节点
	vector<vector<bool>> secondborder(n, vector<bool>(m, false));

	for (int i = 0; i < n; i++)
	{
		//从最左边开始，逆序遍历
		dfs(grid, firstborder, i, 0);
		//从最右边开始，逆序遍历
		dfs(grid, secondborder, i, m-1);
	}
	for (int i = 0; i < m; i++)
	{
		//从最上边开始，逆序遍历
		dfs(grid, firstborder, 0, i);
		//从最下边开始，逆序遍历
		dfs(grid, secondborder, n - 1, i);
	}

	//打印结果
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			if(firstborder[i][j]&&secondborder[i][j])
			{
				cout << i << " " << j << endl;
			}
		}
	}

	
}