算法训练1-day44-图论

眼光中的泪
19 阅读5分钟
  1. 101. 孤岛的总面积 注意设置visited和sum的位置，在dfs的for循环之前就不用在main函数中进入dfs时设置
#include <iostream>
#include <list>
#include <queue>
#include <vector>

using namespace std;

int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; // 上下左右
void dfs(vector<vector<int>> &grid, vector<vector<bool>> &visited, int i, int j,
         int &sum) {
  int n = grid.size();
  int m = grid[0].size();
  visited[i][j] = true;
  sum++;
  for (auto d : dir) {
    int next_i = i + d[0];
    int next_j = j + d[1];
    if (next_i < 0 || next_i >= n || next_j < 0 || next_j >= m) {
      continue;
    }
    if (!visited[next_i][next_j] && grid[next_i][next_j] == 1) {
      dfs(grid, visited, next_i, next_j, sum);
    }
  }
}

int main() {
  int n = 0, m = 0;
  cin >> n >> m;
  vector<vector<int>> grid(n, vector<int>(m, 0));

  int value = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> value;
      grid[i][j] = value;
    }
  }
  int result = 0;
  vector<vector<bool>> visited(n, vector<bool>(m, false));

  for (int i = 0; i < n; i++) {
    if (!visited[i][0] && grid[i][0] == 1) {
      dfs(grid, visited, i, 0, result);
    }
    if (!visited[i][m - 1] && grid[i][m - 1] == 1) {
      dfs(grid, visited, i, m - 1, result);
    }
  }
  for (int j = 0; j < m; j++) {
    if (!visited[0][j] && grid[0][j] == 1) {
      dfs(grid, visited, 0, j, result);
    }
    if (!visited[n - 1][j] && grid[n - 1][j] == 1) {
      dfs(grid, visited, n - 1, j, result);
    }
  }

  result = 0;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j < m; j++) {
      if (!visited[i][j] && grid[i][j] == 1) {
        dfs(grid, visited, i, j, result);
      }
    }
  }
  cout << result << endl; // 输出岛屿数量
  return 0;
}
  1. [102. 沉没孤岛]](kamacoder.com/problempage…) 和101. 孤岛的总面积类似，不过我们可以将边缘岛屿赋特殊值，然后再将特殊值改回1，将1变为0，这样就相当于将孤岛”沉没“了，而不必再设置visited数组
#include<iostream>
#include<vector>
#include<list>
#include<queue>

using namespace std;

int dir[4][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; // 上下左右
void dfs(vector<vector<int>>& grid, int x, int y)
{
	grid[x][y] = 2;
	for (int i = 0; i < 4; i++) {
		int next_x = x + dir[i][0];
		int next_y = y + dir[i][1];
		if (next_x < 0 || next_x >= grid.size() || next_y < 0 || next_y >= grid[0].size()) {
			continue; // 越界
		}
		if (grid[next_x][next_y] == 0 || grid[next_x][next_y] == 2) {
			continue; // 已访问或不是陆地
		}
		dfs(grid, next_x, next_y); // 递归访问相邻陆地
	}
}

int main()
{
	int n = 0, m = 0;
	cin >> n >> m;
	vector<vector<int>> grid(n, vector<int>(m, 0));

	int value = 0;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			cin >> value;
			grid[i][j] = value;
		}
	}
	int result = 0;
	vector<vector<bool>> visited(n, vector<bool>(m, false));

	//深度优先搜索
	for (int i = 0; i < n; i++) {
		if (grid[i][0] == 1) {
			dfs(grid, i, 0); // 从当前陆地开始深度优先搜索
		}
		if (grid[i][m - 1] == 1) {
			dfs(grid, i, m - 1); // 从当前陆地开始深度优先搜索
		}
	}
	//左右
	for (int j = 0; j < m; j++) {
		if (grid[0][j] == 1) {
			dfs(grid, 0, j); // 从当前陆地开始深度优先搜索
		}
		if (grid[n - 1][j] == 1) {
			dfs(grid, n - 1, j); // 从当前陆地开始深度优先搜索
		}
	}

	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;
	}
	return 0;
}
  1. [103. 高山流水]](kamacoder.com/problempage…) 可以反过来想，不是找能符合要求的点，而是从边界开始，看逆过来能走到那些节点，两边都能走到的节点就是符合要求的点
#include <iostream>
#include <list>
#include <queue>
#include <vector>

using namespace std;

int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; // 上下左右
void dfs(vector<vector<int>> &grid, vector<vector<bool>>& visited, int x,
         int y) {
  if (visited[x][y])
    return;
  visited[x][y] = true;
  for (int i = 0; i < 4; i++) {
    int next_x = x + dir[i][0];
    int next_y = y + dir[i][1];
    if (next_x < 0 || next_x >= grid.size() || next_y < 0 ||
        next_y >= grid[0].size()) {
      continue; // 越界
    }
    if (grid[x][y] <= grid[next_x][next_y] && !visited[next_x][next_y]) {
      dfs(grid, visited, next_x, next_y); // 递归访问相邻陆地
    }
  }
}

int main() {
  int n = 0, m = 0;
  cin >> n >> m;
  vector<vector<int>> grid(n, vector<int>(m, 0));

  int value = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> value;
      grid[i][j] = value;
    }
  }
  int result = 0;
  vector<vector<bool>> firstVisited(n, vector<bool>(m, false));
  vector<vector<bool>> secondVisited(n, vector<bool>(m, false));

  for (int i = 0; i < n; i++) {
    dfs(grid, firstVisited, i, 0);
    dfs(grid, secondVisited, i, m - 1);
  }
  for (int j = 0; j < m; j++) {
    dfs(grid, firstVisited, 0, j);
    dfs(grid, secondVisited, n - 1, j);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (firstVisited[i][j] && secondVisited[i][j]) {
        cout << i << " " << j << endl;
      }
    }
  }
  return 0;
}
  1. 104. 建造最大岛屿 先遍历岛屿保存，再检查每一个海水点的上下左右是否有未加入的岛屿
#include <iostream>
#include <vector>
#include <unordered_set>
#include <unordered_map>
using namespace std;
int n, m;
int count;

int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
void dfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y, int mark) {
    if (visited[x][y] || grid[x][y] == 0) return; // 终止条件：访问过的节点 或者 遇到海水
    visited[x][y] = true; // 标记访问过
    grid[x][y] = mark; // 给陆地标记新标签
    count++;
    for (int i = 0; i < 4; i++) {
        int nextx = x + dir[i][0];
        int nexty = y + dir[i][1];
        if (nextx < 0 || nextx >= n || nexty < 0 || nexty >= m) continue;  // 越界了，直接跳过
        dfs(grid, visited, nextx, nexty, mark);
    }
}

int main() {
    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];
        }
    }
    vector<vector<bool>> visited(n, vector<bool>(m, false)); // 标记访问过的点
    unordered_map<int ,int> gridNum;
    int mark = 2; // 记录每个岛屿的编号
    bool isAllGrid = true; // 标记是否整个地图都是陆地
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (grid[i][j] == 0) isAllGrid = false;
            if (!visited[i][j] && grid[i][j] == 1) {
                count = 0;
                dfs(grid, visited, i, j, mark); // 将与其链接的陆地都标记上 true
                gridNum[mark] = count; // 记录每一个岛屿的面积
                mark++; // 记录下一个岛屿编号
            }
        }
    }
    if (isAllGrid) {
        cout << n * m << endl; // 如果都是陆地，返回全面积
        return 0; // 结束程序
    }

    // 以下逻辑是根据添加陆地的位置，计算周边岛屿面积之和
    int result = 0; // 记录最后结果
    unordered_set<int> visitedGrid; // 标记访问过的岛屿
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            count = 1; // 记录连接之后的岛屿数量
            visitedGrid.clear(); // 每次使用时，清空
            if (grid[i][j] == 0) {
                for (int k = 0; k < 4; k++) {
                    int neari = i + dir[k][1]; // 计算相邻坐标
                    int nearj = j + dir[k][0];
                    if (neari < 0 || neari >= n || nearj < 0 || nearj >= m) continue;
                    if (visitedGrid.count(grid[neari][nearj])) continue; // 添加过的岛屿不要重复添加
                    // 把相邻四面的岛屿数量加起来
                    count += gridNum[grid[neari][nearj]];
                    visitedGrid.insert(grid[neari][nearj]); // 标记该岛屿已经添加过
                }
            }
            result = max(result, count);
        }
    }
    cout << result << endl;

}
```# 算法训练1-day43-图论
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