【代码随想录算法训练Day58】卡码网101.孤岛的总面积、卡码网102.沉没孤岛、卡码网103.水流问题、卡码网104.建造最大岛屿

Day58 图论第三天

卡码网101.孤岛的总面积

#include <iostream>
#include <vector>
using namespace std;
int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1}; // 保存四个方向
int count; // 统计符合题目要求的陆地空格数量
void dfs(vector<vector<int>>& grid, int x, int y) {
    grid[x][y] = 0;
    count++;
    for (int i = 0; i < 4; i++) { // 向四个方向遍历
        int nextx = x + dir[i][0];
        int nexty = y + dir[i][1];
        // 超过边界
        if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;
        // 不符合条件，不继续遍历
        if (grid[nextx][nexty] == 0) continue;

        dfs (grid, nextx, nexty);
    }
    return;
}

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];
        }
    }

    // 从左侧边，和右侧边 向中间遍历
    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);
    }
    // 从上边和下边 向中间遍历
    for (int j = 0; j < m; j++) {
        if (grid[0][j] == 1) dfs(grid, 0, j);
        if (grid[n - 1][j] == 1) dfs(grid, n - 1, j);
    }
    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;
}

卡码网102.沉没孤岛

#include <iostream>
#include <vector>
using namespace std;
int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1}; // 保存四个方向
void dfs(vector<vector<int>>& grid, int x, int y) {
    grid[x][y] = 2;
    for (int i = 0; i < 4; i++) { // 向四个方向遍历
        int nextx = x + dir[i][0];
        int nexty = y + dir[i][1];
        // 超过边界
        if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;
        // 不符合条件，不继续遍历
        if (grid[nextx][nexty] == 0 || grid[nextx][nexty] == 2) continue;
        dfs (grid, nextx, nexty);
    }
    return;
}

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];
        }
    }

    // 步骤一：
    // 从左侧边，和右侧边 向中间遍历
    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);
    }

    // 从上边和下边 向中间遍历
    for (int j = 0; j < m; j++) {
        if (grid[0][j] == 1) dfs(grid, 0, j);
        if (grid[n - 1][j] == 1) dfs(grid, n - 1, j);
    }
    // 步骤二、步骤三
    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;
    }
}

卡码网103.水流问题

#include <iostream>
#include <vector>
using namespace std;
int n, m;
int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1};

// 从 x，y 出发 把可以走的地方都标记上
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 nextx = x + dir[i][0];
        int nexty = y + dir[i][1];
        if (nextx < 0 || nextx >= n || nexty < 0 || nexty >= m) continue;
        if (grid[x][y] < grid[nextx][nexty]) continue; // 高度不合适

        dfs (grid, visited, nextx, nexty);
    }
    return;
}
bool isResult(vector<vector<int>>& grid, int x, int y) {
    vector<vector<bool>> visited(n, vector<bool>(m, false));

    // 深搜，将x,y出发 能到的节点都标记上。
    dfs(grid, visited, x, y);
    bool isFirst = false;
    bool isSecond = false;

    // 以下就是判断x，y出发，是否到达第一组边界和第二组边界
    // 第一边界的上边
    for (int j = 0; j < m; j++) {
        if (visited[0][j]) {
            isFirst = true;
            break;
        }
    }
    // 第一边界的左边
    for (int i = 0; i < n; i++) {
        if (visited[i][0]) {
            isFirst = true;
            break;
        }
    }
    // 第二边界右边
    for (int j = 0; j < m; j++) {
        if (visited[n - 1][j]) {
            isSecond = true;
            break;
        }
    }
    // 第二边界下边
    for (int i = 0; i < n; i++) {
        if (visited[i][m - 1]) {
            isSecond = true;
            break;
        }
    }
    if (isFirst && isSecond) return true;
    return false;
}


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];
        }
    }
    // 遍历每一个点，看是否能同时到达第一组边界和第二组边界
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (isResult(grid, i, j)) {
                cout << i << " " << j << endl;
            }
        }
    }
}

卡码网104.建造最大岛屿

#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;

}